Subjective Topicwise Question Bank ( Q&A )

1.4 Electric Charge and its Basic Properties

SA-I (1-2 Marks)

1. State two basic properties of electric charge.

2. What is meant by conservation of charge?
3. What is the least possible value of charge?
4. What is the cause of quantisation?
5. Define Induction.
6. During lightening you are safer inside a house than under a tree. Why?

7. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar
phenomenon is observed with many other pairs of bodies. Explain how this observation is
consistent with the law of conservation of charge.

1.5 Coulomb's Law

SA-1 (1-2 Marks)
8. Two identical conducting balls A and B have charges -Q and +3Q respectively. They are
brought in contact with each other and then separated by a distance d apart. Find the
nature of the Coulomb force between them.
(2019 Series: BVM/4)
SA-II (3 Marks)
9. State Coulomb's law and write its mathematical expression

10. Deduce Coulomb's law from Gauss's Law.

1.6 Forces between Multiple Charges

SA-I (1-2 Marks)

11. Two point charges ' 𝒒𝟏 ' and ' 𝒒𝟐 ' are placed at a distance ' 𝒅 ' apart as shown in the
figure. The electric field intensity is zero at a point ' 𝐏′ on the line joining them as shown.
Write two conclusions that you can draw from this.
(2014 Comptt. Delhi)

Ans. (i) Two point charges ' 𝑞1 ' and ' 𝑞2 ' should be of opposite nature.
(ii) Magnitude of charge 𝑞1 must be greater than that of charge 𝑞2 .

𝟏
12. Plot a graph showing the variation of coulomb force (𝑭) versus (𝒓𝟐), where 𝒓 is the
distance between the two charges of each pair of charges: ( 𝟏𝝁𝐂, 𝟐𝝁𝐂 ) and ( 𝟐𝝁𝐂, −𝟑𝝁𝐂 ).
Interpret the graphs obtained.
 (2011 All India)

Ans. For (1𝜇C, 2𝜇C)

9×109 (1×10−6)(2×10−6) 0.018

F1 = =
𝑟2 𝑟2

and for (2𝜇C, −3𝜇C)

9×109 (2×10−6)(−3×10−6 ) −0.054

F2 = =
𝑟2 𝑟2

Here positive slope depicts that force is repulsive in nature and negative slope depicts that
the force is attractive in nature.

13. (a) Two spherical conductors of radii 𝑹𝟏 and 𝑹𝟐 (𝑹𝟐 > 𝑹𝟏 ) are charged. If they are
connected by a conducting wire, find out the ratio of the surface charge densities
on them.

(b) A steady current flows in a metallic conductor of non-uniform cross-section.

Which of these quantities is constant along the conductor: current, current
density, electric field, drift speed?
(2015 Comptt. Delhi)

Ans. (a) When two charged spherical conductors of Radii 𝑅1 and 𝑅2 respectively (𝑅2 > 𝑅1 )
are connected by a conducting wire, we know that the common potential (V) is given by,

𝑞1 𝑞2
V= =
𝑐1 𝑐2

∵ C for a spherical conductor, C = 4𝜋𝜖0 R,

𝑞 𝑞
We have, 4𝜋𝜀1R = 4𝜋𝜀2R
0 1 0 1
 𝑞 𝑞 𝑞 R
⇒ R1 = R2 ⇒ 𝑞1 = R1
1 2 2 2
𝜎1 𝑞1 4𝜋𝜀0R22 𝑞
and 𝜎2
= 4𝜋𝜀 2 × … [∴ 𝜎 = 4𝜋𝜀 2
0 R1 𝑞2 0R
𝑞1 R22 R1 R22 R2
= × = × =
𝑞2 R1 2 R2 R21 R1
𝜎1 R2
∴ =
𝜎2 R1

(b) Current

1.7 Electric Field

SA-1 (2 Marks)

14. A small metal sphere carrying charge +𝑸 is located at the centre of a spherical cavity in a
large uncharged metallic spherical shell. Write the charges on the inner and outer surfaces
of the shell. Write the expression for the electric field at the point 𝑷𝟏.

(2014 Comptt. Delhi)

Ans. (i) Charge on inner surface: −𝑄

(ii) Charge on outer surface: +𝑄
1 Q
(iii) Electric field at point P1 (E) = 4𝜋𝜀 𝜋12
0

15. Two point charges +𝒒 and −𝟐𝒒 are placed at the vertices ' 𝑩 ' and ' 𝑪 ' of an equilateral
triangle 𝑨𝑩𝑪 of side ' 𝒂 ' as given in the figure. Obtain the expression for (i) the magnitude
and (ii) the direction of the resultant electric field at the vertex 𝑨 due to these two
charges.

(2014 Comptt. All India)

Ans. (i) Magnitude,

 1 𝑞
|𝐸⃗AB | = =E
4𝜋𝜖0 𝑎 2
1 2𝑞
|𝐸⃗AC | = = 2E
4𝜋𝜖0 𝑎 2
1 1
Enet = √(2E)2 + E2 + 2 × 2E × E (− 2) + ⋯ [∵ cos 120∘ = − 2

= √4E2 + E2 − 2E2
1 𝑞√3
= √3E2 = E√3 =
4𝜋𝜖0 𝑎2

(ii) Direction of resultant electric field at vertex A,

EAB sin 120 ∘

tan 𝛼 = E ∘
AC +EAB cos 120
√3 E×√3
E× √3 1
= 2
−1 = 2
3E = = = tan 30∘
2E×+E×( ) 3 √3
2 2
∴ 𝛼 = 30∘ (with side AC )

16. Two point charges +𝟑𝒒 and −𝟒𝒒 are placed at the vertices ' 𝑩 ' and ' 𝑪 ' of an equilateral
triangle 𝐀𝐁𝐂 of side ' 𝒂 ' as given in the figure.
Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric
field at the vertex A due to these two charges.
(2014 Comptt. All India)
Ans. (i) Magnitude,

1 3𝑞 1 𝑞
⃗⃗⃗⃗⃗⃗⃗
|𝐸 𝐴𝐵 | = = 3E, where [E = 4𝜋𝜖 ]
4𝜋𝜖0 𝑎 2 0 𝑎2
1 4𝑞
⃗⃗⃗⃗⃗⃗
|𝐸 𝐴𝐶 | = = 4E
4𝜋𝜖0 𝑎 2
1
Enet = √(3E)2 + (4E)2 + 2(3E) × (4E) × (− 2)
∵ 𝜃 = 120∘
EAB {cos 𝜃 = − 1}
2

= √9E2 + 16E2 − 12E2

1 𝑞√13
= √13E2 = E√13 =
4𝜋𝜖0 𝑎 2

(ii) Direction,

√3
|𝐸𝐴𝐵 |sin 120 ∘ 3E×
2
tan 𝛼 = |𝐸 ∘ = 1
𝐴𝐶 |+|𝐸𝐴𝐵 |cos 120 4E+3E×−( )
2
3E√3×2 3√3 3√3
tan 𝛼 = 2×5E
= 5
∴ 𝛼 = tan−1 ( 5
)

17. Four point charges 𝑸𝒒, 𝑸 and 𝒒 are placed at the corners of a square of side ' 𝒂 ' as shown
in the figure. Find the resultant electric force on a charge 𝑸. (2018)
Ans. Let us find the force on the charge 𝑄 at the point 𝐶 diagonal

AC = √AB2 + BC2 = √𝑎 2 + 𝑎2 = √2𝑎 2 = √2𝑎

1 𝑞1 𝑞2
We know F = ⋅
4𝜋𝜀0 𝑟

1 (Q×Q) 1 Q2
F1 ( along AC) = 4𝜋𝜀 ⋅ (√2𝑎)2 = 4𝜋𝜀 (2𝑎2) … (𝑖)
0 0
1 𝑞Q
F2 ( along BC) = 4𝜋𝜀 ( 𝑎2 ) … . (𝑖𝑖)
0
1 Q𝑞
F3 ( along DC) = 4𝜋𝜀 ⋅ 𝑎2 … . (𝑖𝑖𝑖)
0

Resultant force F23 due to these two equal forces as per above equations (ii) and (iii).

… [∵ F = √F1 2 + F2 2 + 2 F1 F2 cos 𝜃

1 𝑞Q(√2)
F23 = 4𝜋𝜀 ( ) (𝑎𝑙𝑜𝑛𝑔 𝐴𝐶 )
0 𝑎2

Net force on charge Q = F1 + F23

 1 Q2 1 (𝑞Q√2)
= [4𝜋𝜀 (2𝑎2)] + [4𝜋𝜀 ⋅ 𝑎2
]
0 0
1 Q Q
= 4𝜋𝜀 (𝑎2) [ 2 + √2𝑞]
0

This force is directed towards AC.

18 Three point charges 𝑞, −4𝑞 and 2𝑞 are placed at the vertices of an equilateral triangle ABC
of side ' 𝑙 ' as shown in the figure. Obtain the expression for the magnitude of the resultant
electric force acting on the charge 𝑞.
(2018)

Ans. Force on charge 𝑞 (at point A) due to charge 4𝑞 (at point B)

1 𝑞×4𝑞
F1 = 4𝜋𝜀 ( 𝑙2
)
0
1 4𝑞 2
= ( )
4𝜋𝜀0 𝑙2

…(i) [along AB

Force on charge 𝑞 (at point 𝐴 ) due to charge 2𝑞 (at point C)

1 𝑞×2𝑞 1 2𝑞 2
F2 = ( )= ( )
4𝜋𝜀0 𝑙2 4𝜋𝜀0 𝑙2

...(ii) [along CA

The forces F1 and F2 are inclined to each other at an angle of 120 ∘ .

Hence resultant electric force on charge ' 𝑞 '

F = √F12 + F22 + 2 F1 F2 cos 𝜃 = √F12 + F22 + 2 F1 F2 (cos 120∘ )

1
= √F12 + F22 − F1 F2 … [∵ cos 120∘ = −
2
1 4𝑞 2 2 1 2𝑞 2 2 1 4𝑞 2 1 2𝑞 2
= √[ ( )] + [ ( )] − 2 [ ⋅ × ⋅ ]
4𝜋𝜀0 𝑙2 4𝜋𝜀0 𝑙2 4𝜋𝜀0 𝑙2 4𝜋𝜀0 𝑙2
1 𝑞2 1 𝑞2
= ⋅ √16 + 4 − 8 = 4𝜋𝜀 ⋅ ⋅ √12
4𝜋𝜀0 𝑙2 0 𝑙2
1 𝑞2 1 2√3𝑞 2
=
4𝜋𝜀0
⋅
𝑙2
√3 × 4 = 4𝜋𝜀 ( 𝑙2
)
0
19. Two point charges 4𝜇C and +1𝜇C are separated by a distance of 2 m in air. Find the point on
the linejoining charges at which the net electric field of the system is zero.
(2017 Comptt. Outside Delhi)
Ans. Given : 𝑞1 = 4𝜇C, 𝑞2 = 1𝜇C, 𝑟 = 2 m
Let us say that at a point P (distance 𝑥 ) the net field is zero.

1 4
E1 = … . (𝑖)
4𝜋𝜖0 𝑥 2
1 4
E2 = … . (𝑖𝑖)
4𝜋𝜖0 (2 − 𝑥)2
E1 = E2

4 1
∴ 2
= … . [𝐹𝑟𝑜𝑚 (𝑖 ) & (𝑖𝑖)
𝑥 (2 − 𝑥)2
2 2 1 2
( ) =( )
𝑥 (2 − 𝑥)

Taking square root on both sides,

2 1
⇒ 𝑥
= 2−𝑥 ⇒ 𝑥 = 4 − 2𝑥
4
⇒ 𝑥=3 m

At this point, the net electric field of the system is zero.

1.8 Electric Field Lines

SA-I (1-2 Marks)

20. Why must electrostatic field be normal to the surface at every point of a charged
conductor? (2012 Delhi)

Ans. So that tangent on charged conductor gives the direction of the electric field at that point.

21. Why do the electric field lines not form closed loops?
(2012 Comptt. All India)

Ans. Electric field lines do not form closed loops because the direction of an electric field is from
positive to negative charge. So one can regard a line of force starting from a positive charge
and ending on a negative charge. This indicates that electric field lines do not form closed
loops.

22. Why do the electric field lines never cross each other?
(2014 All India)

Ans. The electric lines of force give the direction of the electric field. In case, two lines of force
intersect, there will be two directions of the electric field at the point of intersection, which
 is not possible.

23. A point charge +𝑸 is placed in the vicinity of a conducting surface. Draw the electric field
lines between the surface and the charge.
(2016 Comptt. Outside Delhi, 2019 Series: BVM/1)

Ans. Electric field lines between surface and charge.

24. Draw the pattern of electric field lines, when a point charge −𝑸 is kept near an uncharged
conducting plate.
(2019 Series: BVM/1)
Ans.

1.9 Electric Flu0078

SA-I (1-2 Marks)

25. A charge ' 𝒒 ' is placed at the centre of a cube of side 𝒍. What is the electric flux passing
through each face of the cube?
(2012 All India)

Ans. Electric flux through each phase of the cube

1 1 𝑞 𝑞
= 6 𝜙𝐸 = 6 𝜀 = 6𝜀
0 0
26. A charge ' 𝒒 ' is placed at the centre of a cube of side 𝒍. What is the electric flux passing
through two opposite faces of the cube?
(2012 All India)
𝑞
Ans. 𝜙𝐸 = 3𝜀
0

27. ⃗⃗ = 𝟑 × 𝟏𝟎𝟑 𝒊ˆ N/C through a square of side 𝟏𝟎 𝐜𝐦,

What is the flux due to electric field 𝑬
⃗ ?
when it is held normal to 𝐄

(2015 Comptt. All India)

Ans. ⃗ = 3 × 103 𝑖ˆ N/C

Given: E

10 10
A = 10 × 10 cm2 = × m2
100 100

𝜙 = 𝐸⃗ × 𝐴 = 𝐸𝐴cos 𝜃 − [∵ 𝜃 = 0 and cos 𝜃 = 1

= 𝐸𝐴
10 10
= (3 × 103 ) × ( × ) = 30Nm2 C −1
100 100

28. Consider a uniform electric field 𝐸⃗ = 3 × 103 𝑖ˆ N/C. Calculate the flux of this field through a
square surface of area 10 cm2 when
(i) its plane is parallel to the 𝑦 − 𝑧 plane, and
(ii) the normal to its plane makes a 60∘ angle with the 𝑥-axis.
(2013 Comptt. Delhi)

Ans. Given: ⃗E = 3 × 103 𝑖ˆ

10
S = 10 cm2 = 102 ×102 m2 = 10−3 m2

(i) Flux (𝜙) = E × S = 3 × 103 × 10−3 = 3Nm2 C−1

1
(ii) (3 × 103 ) × (10−3 )cos(60∘ ) = 3 × = 1.5Nm2 C−1
2

29. Given a uniform electric field 𝐸⃗ = 5 × 103 𝑖ˆ N/C, find the flux of this field through a square of
10 cm on a side whose plane is parallel to the 𝑦 − 𝑧 plane. What would be the flux through
the same square if the plane makes a 30∘ angle with the 𝑥-axis?
(2014 Delhi)
Ans. Given: 𝐸⃗ = 5 × 103 𝑖 N/C, A = 10 × 10 × 10−4 m2 Flux (𝜙) = EAcos 𝜃
(i) For first case, 𝜃 = 0, cos 0 = 1
∴ Flux = (5 × 103 ) × (10 × 10 × 10−4 ) = 50Nm2 C−1
(ii) Angle of square plane with 𝑥-axis = 30∘
Hence the 𝜃 will be 90∘ − 30∘ = 60∘

∴ EA cos 𝜃 = (5 × 103 ) × (10 × 10 × 10−4 ) × cos 60

1
= 50 × 2 = 25Nm2 C−1
30. ⃗ = 𝟐 × 𝟏𝟎𝟑𝒊ˆ 𝐍/𝐂, find the flux of this field through a square
Given a uniform electric field 𝐄
of side 𝟐𝟎 𝐜𝐦, whose plane is parallel to the 𝒚 − 𝒛 plane. What would be the flux through
the same square, if the plane makes an angle of 𝟑𝟎∘ with the 𝒙-axis?
(2014 Delhi)
Ans. Similar to Queston 27.
Hint:
(i) 80Nm2 C−1
(ii) 40Nm2 C −1

31. Given a uniform electric field ⃗𝑬

⃗ = 𝟒 × 𝟏𝟎𝟑 𝒊ˆ N/C. Find the flux of this field through a square
of 𝟓 𝐜𝐦 on a side whose plane is parallel to the 𝐘 − 𝐙 plane. What would be the flux
through the same square if the plane makes a 𝟑𝟎∘ angle with the 𝒙-axis?

Ans. Similar to Question 27.

Hint: (i) 10Nm2 C −1
(ii) 5Nm2 C−1

LA (5 Marks)

32. (a) Define electric flux. Write its SI units.

(b) The electric field components due to a charge inside the cube of side 𝟎. 𝟏 𝐦 are as
shown:

𝐸𝑥 = 𝛼𝑥
where 𝛼 = 500NC−𝑚
E𝑦 = 0, 𝐸𝑧 = 0.

Calculate:
(i) the flux through the cube, and
(ii) the charge inside 𝑥 the cube.
(2008 All India)

Ans. (a) Electric flux through a surface represents the total number of electric lines of force
crossing the surface.
∴ S.I. unit is Nm2 C −1 .

(b) (i) Flux through R.H.S. of the cube is

 𝜙1 = E𝑥1 ⋅ A = (𝛼𝑥) ⋅ (𝑙)2 ∵ 𝜃 = 0∘
−𝑚
= (500 × 0.2)(0.1)2 [𝛼 = 500NC
𝑥 = 0.2 m
= 1Nm2 C−1 𝑙 = 0.1 m

(ii) Flux through L.H.S. of the cube is

𝜙2 = E𝑥2 ⋅ A = −(𝛼𝑥) ⋅ 𝑙 2
= −(500 × 0.1)(0.1)2 = −0.5Nm2 /C

Net flux 𝜙 = 𝜙1 + 𝜙2 = 1 − 0.5 = 0.5Nm2 /C

𝑞
𝜙 =𝜀 ∴ 𝑞 = 𝜀0 𝜙
(iii) As 0
⇒𝑞 = 8.854 × 10−12 × 0.5 = 4.4 × 10−12 C

33 (a) Define electric flux. Is it a scalar or a vector quantity? A point charge 𝒒 is at a distance
𝒅
of 𝟐 directly above the centre of a square of side 𝒅, as shown in the figure. Use Gauss' law
to obtain the expression for the electric flux through the square.
(b) If the point charge is now moved to a distance ' 𝒅 ' from the centre of the square and
the side of the square is doubled, explain how the electric flux will be affected.
(2018)

Ans. (a) Electric flux through a given surface is defined as the dot product of electric field and
area vector over that surface.
⃗⃗⃗⃗
Alternatively 𝜙 = ∫ 𝐸⃗ ⋅ 𝑑𝑆
𝑠

It is a scalar quantity.
Constructing a cube of side ' 𝑑 ' so that charge ' 𝑞 ' gets placed within this cube (Gaussian
surface).
According to Gauss's law,

charge enclosed 𝑞
Electric flux 𝜙 = 𝜀0
=𝜀
0
 This is the total flux through all the six faces of the cube.
Hence electric flux through the square

1 𝑞 𝑞
= × =
6 𝜀0 6𝜀0

(b) If the charge is moved to a distance 𝑑 and the side of the square is doubled, the cube will be
constructed to have a side 2𝑑 but the total charge enclosed in it will remain the same.
Hence the total flux through the cube and therefore the flux through the square will remain
the same as before.

1.10 Electric Dipole

SA-I (1-2 Marks)

34. Define electric dipole moment. Write its S.I. unit.

(2011 All India)

Ans. Electric dipole moment of an electric dipole is defined as the product of the magnitude of
either charge and dipole length.

In other words, dipole moment may be defined as the torque acting on an electric dipole,
placed perpendicular to a uniform electric field of unit strength.
or Strength of electric dipole is called dipole moment.

𝑝 = 𝑞(2𝑙 )
 S.I. unit of dipole (𝑝) is coulomb metre ( Cm ).

35. Is the electric field due to a charge configuration with total charge zero, necessarily zero?
Justify.
(2012 Comptt. All Delhi)

Ans. No, it is not necessarily zero. If the electric field due to a charge configuration with total
charge is zero because the electric field due to an electric dipole is non-zero.

36. What is the electric flux through a cube of side 𝟏 𝐜𝐦 which encloses an electric dipole?
(2015 Delhi)

Ans. Zero because the net charge of an electric dipole ( +𝑞 and −𝑞 ) is zero.

1.11 Dipole in a Uniform External Field

SA-1 (1-2 Marks)

37. Which orientation of an electric dipole in a uniform electric field would correspond to
stable equilibrium?
(2008 All India)

Ans. When dipole moment vector is parallel to electric field vector

⃗P ∥ ⃗E

38. In which orientation, a dipole placed in a uniform electric field is in (i) stable, (ii) unstable
equilibrium?
(2010 Delhi)
Ans. (i) For stable equilibrium, a dipole is placed parallel to the electric field.
(ii) For unstable equilibrium, a dipole is placed antiparallel to the electric field.

39. Write the expression for the work done on an electric dipole of dipole moment 𝒑 ⃗ in
turning it from its position of stable equilibrium to a position of unstable equilibrium in a
uniform electric field ⃗𝑬
⃗.
(2013 Comptt. Delhi)

Ans. Torque, acting on the dipole is, 𝜏 = 𝑝Esin 𝜃

𝜃
𝜏 = ∫𝜃 2 𝑝 Esin 𝜃𝑑𝜃 ⇒ 𝜏 = 𝑝E[cos 𝜃1 − cos 𝜃2 ]
1
∴ 𝜏 = 𝑝E[cos 0∘ − cos 180 ∘ ]
= 𝑝E[1 − (−1)] = 2𝑝E ∴ 𝜏 = 2𝑝E

40. What will happen if the field were not uniform?

(2017 Delhi)
Ans. If the electric field is non uniform, the dipole experiences a translatory force as well as a
torque.
41. Derive an expression for the torque experienced by an electric dipole kept in a uniform
electric field.
(2008 Delhi)

Ans. Consider an electric dipole consisting of charges +𝑞 and −𝑞 and of length 2𝑎 placed in a
⃗ making an angle 𝜃 with it. It has a dipole moment of magnitude,
uniform electric field E

𝑝 = 𝑞 × 2𝑎

Force exerted on charge +𝑞 by field,

⃗ = 𝑞E
F ⃗ (along E
⃗ )

Force exerted on charge −𝑞 by field,

⃗F = 𝑞E⃗ (opposite to ⃗E )
∴ ⃗Ftotal = +𝑞E
⃗ − 𝑞E ⃗ =0

Hence the net translating force on a dipole in a uniform electric field is zero. But the two
equal and opposite forces act at different points of the dipole. They form a couple which
exerts a torque.
Torque = Either force x
Perpendicular distance between the two forces

𝜏 = 𝑞E × 2𝑎sin 𝜃
𝜏 = 𝑝 Esin 𝜃

⋯ [∵ 𝑝 = 𝑞 × 2𝑎; 𝑝 is dipole moment

As the direction of torque 𝜏 is perpendicular to both 𝑝 and ⃗E, so we can write 𝜏 = 𝑝 × ⃗E.

42. ⃗⃗ with its dipole moment 𝒑

An electric dipole is placed in a uniform electric field 𝑬 ⃗ parallel
to the field. Find
(i) the work done in turning the dipole till its dipole moment points in the direction
opposite to ⃗𝑬⃗.
 (ii) the orientation of the dipole for which the torque acting on it becomes maximum.
(2014 Comptt. All India)

Ans.

𝜃 𝜋
(i) W = ∫𝜃 2 𝜏𝑑𝜃 ⇒ ∫0 𝑝𝐸sin 𝜃𝑑𝜃
1
⇒ 𝑝𝐸[−cos 𝜃] ⇒ 𝑝𝐸[cos 𝜋 − cos 0]
⇒ 𝑝𝐸[(−1) − (1)] ⇒ −2𝑝E

(ii) 𝜏 = 𝑝 × ⃗E = 𝑝Esin 𝜃
𝜋
For 𝜃 = 2 , sin 𝜃 = 1 and 𝜏 is maximum.

43. Depict the orientation of the dipole in (2017 Delhi)

(a) stable,
(b) unstable equilibrium in a uniform electric field.
Ans. (a) For stable equilibrium, the angle between 𝑝 and E is 0∘ ,

(b) For unstable equilibrium, the angle between 𝑝 and E is 180∘ ,

44. An electric dipole of dipole moment (𝒑 ⃗ ) is kept in a uniform electric field ⃗𝑬

⃗ . Show
graphically the variation of torque acting on the dipole ( 𝐫 ) with its orientation ( 𝜽 ) in the
field. Find the orientation in which torque is
(i) zero and (ii) maximum.
(2023 Series: GEFH1/5)
Ans. (i) Torque is zero for orientation corresponding to 𝜃 = 0∘ and 𝜃 = 180∘ .

𝜋 𝜋
(ii) Torque is maximum for orientation corresponding to 𝜃 = 2 and 3 2 .

45. Identify two pairs of perpendicular vectors in the expression.

(2015 Comptt. Delhi)

Ans. Two pairs of perpendicular vectors are:

(a) 𝜏 is perpendicular to 𝑝
⃗
(b) 𝜏 is perpendicular to E

SA-II (3 Marks)

46. (a) Draw a graph of E versus 𝒓 for 𝒓 ≫ 𝒂.

(b) If this dipole were kept in a uniform external electric field 𝐄𝟎 , diagrammatically
represent the position of the dipole in stable and unstable equilibrium and write the
expressions for the torque acting on the dipole in both the cases.
(2017 Outside Delhi)

Ans. (a) Graph between E vs. 𝑟

(b) (i) Diagrammatic representation
(ii) Torque acting on these cases

 In stable equilibrium, torque is zero (𝜃 = 0)

 In unstable equilibrium also, torque is zero (𝜃 = 180∘ )

LA (5 Marks)

47. Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the
electric field of a dipole at a point on the equatorial plane of the dipole.
(2013 All India, 2019 Series: BVM/1)

𝐎𝐫
Derive the expression for electric field at a point on the equatorial line of an electric
dipole. (2017 Delhi)

Ans. Electric dipole moment: It is the product of the magnitude of either charge and distance
between them.

𝑞 = 𝑞 × 2𝑙

It is a vector quantity whose direction is from negative to positive charge.

Expression:
Electric field intensity at 𝑃 due to +𝑞 charge is

1 𝑞
𝐸⃗+ = along PD
4𝜋𝜀0 BP2
1 𝑞
= 4𝜋𝜀0 (𝑟2 +𝑙 2)
along PD ….(i)

Electric field intensity at P due to −𝑞 charge is,

1 𝑞
⃗⃗⃗⃗
𝐸− = along PA
4𝜋𝜀0 AP2
1 𝑞
= 4𝜋𝜀0 (𝑟2 +𝑙 2)
along PA … . (ii)

1 𝑞
From (i) and (ii), |𝐸⃗+ | = |𝐸⃗− | = 4𝜋𝜀 (𝑟2 +𝑙 2)
…..(iii)
0

Net electric field intensity due to the electric dipole at point 𝑃.

2 2 + 2E E cos 2𝜃
∴ E = √E+ + E− + −
2 2 + 2E2 cos 2𝜃
⇒ E = √E+ + E− + … [∵ E− = E+
√2E+2 2
⇒ E= + 2E+ cos 2𝜃
√2E+2
⇒ E= (1 + cos 2𝜃)
2
√2E+ 2cos2 𝜃
⇒ E=
1 𝑞
∴ E= 2E+ cos 𝜃 = 2 × 4𝜋𝜀 (𝑟2 +𝑙2 ) cos 𝜃
0

...[Using equation (iii)

1
Now from △ OAP, cos 𝜃 = √𝑟2
+𝑙 2
 1 𝑞 𝑙
E = 2 × 4𝜋𝜀 (𝑟 2 +𝑙 2 ) × (𝑟2 +𝑙2)1/2
0
𝑞×2𝑙
⇒E = 4𝜋𝜀0(𝑟2 +𝑙 2)3/2

Since 𝑞 × 2𝑙 = 𝑝
...[ 𝒑 is dipole moment
𝑝
E= along (−)𝑥-axis
4𝜋𝜀0(𝑟2 +𝑙 2 )3/2

If 𝑙 ≪ 𝑟 i.e., dipole is short, then 𝑙 2 can be neglected as compared to 𝑟 2 .

𝑝
Hence E = 4𝜋𝜀 𝑟 3 along (−)𝑥-axis
0

48 (a) Deduce the expression for the torque acting on a dipole of dipole moment 𝒑⃗ in the
⃗⃗ .
presence of a uniform electric field 𝑬
Obtain the expression for the torque experienced by an electric dipole of dipole moment
in a uniform electric.
Or
Derive the expression for the torque acting on an electric dipole, when it is held in a
uniform electric field. Identify the orientation of the dipole in the electric field, in which it
attains a stable equilibrium.
(2020 Series: 𝐇𝐌𝐉/𝟓 )

(b) Consider two hollow concentric spheres, 𝐒𝟏 and 𝑺𝟐 , enclosing charges 𝟐𝑸 and 𝟒𝑸
respectively as shown in the figure.

(i) Find out the ratio of the electric flux through them
(ii) How will the electric flux through the sphere 𝑺𝟏 change if a medium of dielectric
constant ′ 𝜺′𝒓 is introduced in the space inside 𝑺𝟏 in place of air? Deduce the necessary
expression.
(2014 All India)

Ans. (a) Torque on electric dipole. Consider an electric dipole consisting of two equal and
opposite point charges separated by a small distance 2𝑎 having dipole moment

|𝑝| = 𝑞(2𝑎 )
 ⃗ at an angle 𝜃.
Let the dipole held in a uniform external electric field E
∴ Force on charge (+𝑞) = 𝑞E ⃗ along the direction of 𝐸⃗

Force on charge (−𝑞) = −𝑞𝐸⃗ along the opposite direction of 𝐸⃗

∴ Net translatory force on the dipole

⃗ − 𝑞E
= 𝑞E ⃗ =0

So net force on the dipole is zero.

Since 𝐸⃗ is uniform, hence the dipole does not undergo any translatory motion.
These forces being equal, unlike and parallel, form a couple, which rotates the dipole in
clock-wise direction
∴ Magnitude of torque = Force × arm of couple 𝜏 = F. AC = 𝑞E. ABsin 𝜃 = (𝑞E)2𝑎sin 𝜃
or 𝜏 = 𝑞(2𝑎)Esin 𝜃

or 𝜏 = 𝑝ESin𝜃
∴ 𝜏 = 𝑝 × ⃗E

⃗ 𝑎)
… [∵ 𝑝 = 𝑞(2

⃗ is given by right hand screw rule and is normal to 𝒑

[The direction of 𝝉 ⃗.
⃗ and 𝐄

Special cases
(i) When 𝜃 = 0 then 𝜏 = PEsin 𝜃 = 0
∴ Torque is zero and the dipole is in stable equilibrium.
(ii) When 𝜃 = 90∘ then 𝜏 = PEsin 90∘ = PE
∴ The torque is maximum
(b) (i) Ratio of flux:
𝑄
We know electric flux (𝜙) = 𝜀
0
2𝑄
Thus, 𝜙1 due to 𝑆1 =
𝜀0
2Q+4Q 6Q
𝜙2 due to S2 = 𝜀0
= 𝜀0
𝜙2 6𝑄/𝜀0 3
𝜙1
= 2𝑄/𝜀 = 1 ∴ Ratio = 𝟑: 𝟏
0

2Q 1
(ii) 𝜙m = 𝜀0
×𝜀
𝑟
∴ Electricflux through the sphere 𝑆1 decreases with the introduction of dielectric inside it.

49. (a) An electric dipole of dipole moment ⃗𝒑 consists of point charges +𝒒 and −𝒒 separated
by a distance 𝟐𝒂 apart. Deduce the expression for the electric field 𝐄 ⃗ due to the dipole at a
distance 𝒙 from the centre of the dipole on its axial line in terms of the dipole moment 𝒑 ⃗.
⃗ ⃗
Hence show that in the limit 𝒙 ≫ 𝒂, 𝐄 ⟶ 𝟐𝒑/(𝟒𝝅𝜺𝟎 𝒙 𝟑)

⃗ = 𝟐𝒙𝒊ˆ, find the net electric flux through the

(b) Given the electric field in the region 𝐄
cube and the charge enclosed by it.
(2015 Delhi)

Ans. (a) Expression for magnetic field due to dipole on its axial lane:

Electric field intensity at point 𝑝 due to charge −𝑞,

1 𝑞
⃗⃗⃗⃗⃗⃗⃗
𝐸 −𝑞 = 4𝜋𝜀 ⋅ (𝑥+𝑎)2 (𝑥ˆ)
0
 Due to charge +𝑞,

1 𝑞
⃗⃗⃗⃗⃗⃗⃗
𝐸 +𝑞 = 4𝜋𝜀 ⋅ (𝑥−𝑎)2 (𝑥ˆ)
0

Net Electric field at point P, ⃗E = ⃗E−𝑞 + ⃗E+𝑞

𝑞 1 1
= 4𝜋𝜀 × [(𝑥−𝑎)2 − (𝑥+𝑎)2] (𝑥ˆ)
0
1 4𝑎𝑞𝑥 1 (𝑞×2𝑎)2𝑥
= 4𝜋𝜀 [(𝑥 2−𝑎2)2] (𝑥ˆ) = 4𝜋𝜀 2 2 2 (𝑥
ˆ)
0 0 (𝑥 −𝑎 )
⃗ = 1 ⋅ 22𝑝𝑥2 2 𝑥ˆ ∴ 𝑝 = (𝑞 × 2𝑎)
E 4𝜋𝜀 (𝑥 −𝑎 ) 0

For 𝑥 ≫ 𝑎

1 2𝑝
(𝑥 2 − 𝑎2 )2 ≃ 𝑥 4 ⃗ =
∴𝐄 ⋅ 𝑥 3 𝑥ˆ
4𝜋𝜀 0

(b) Only the faces perpendicular to the direction of 𝑥-axis, contribute to the Electric flux. The
remaining faces of the cube give zero Total flux 𝜙 = 𝜙1 + 𝜙II

∮ ⃗E ⋅ ⃗⃗⃗⃗
𝑑𝑠 + ⃗E ⋅ ⃗⃗⃗⃗
𝑑𝑠 = 0 + 2(𝑎) ⋅ 𝑎 2
I II
∴𝜙 = 2𝑎3

𝑞
Charge enclosed (𝑞) = 𝜙𝜖0 = 2𝑎 3 𝜖0 [∵ 𝜙 =
𝜖0

1.12 Gauss's Law

SA-I (1-2 Marks)

50. If the radius of the Gaussian surface enclosing a charge is halved, how does the electric
flux through the Gaussian surface change?
(2008 All India)
 Q
Ans. Electric flux 𝜙E is given by 𝜙E = ∮ ⃗E ⋅ 𝑑𝑠 = 𝜀
0

...where [𝑄 is total charge inside the closed surface ∴ On changing the radius of sphere, the
electric flux through the Gaussian surface remains same.

51. Figure shows three point charges, +𝟐𝒒, −𝒒 and +𝟑𝒒. Two charges +𝟐𝒒 and −𝒒 are
enclosed within a surface ' 𝑺 '. What is the electric flux due to this configuration through
the surface ' 𝐒 '?
(2010 Delhi)

Ans. ⃗ ⋅ ⃗⃗⃗⃗⃗
Electric flux = ∮ E 𝑑S
𝑞
⃗ ⋅ ⃗⃗⃗⃗⃗
According to Gauss's law, 𝜙 = ∮ E 𝑑S= 1
s 𝜀0
… 𝒘𝒉𝒆𝒓𝒆[𝒒𝟏 is the total charge enclosed by the surface 𝑺

2𝑞−𝑞 𝑞 𝒒
𝜙= = ∴ Electric flux, 𝝓 =
𝜀0 𝜀0 𝜺𝟎

52. Two charges of magnitudes −𝟐𝑸 and +𝑸 are located at points (𝒂, 𝟎) and (𝟒𝒂, 𝟎)
respectively. What is the electric flux due to these charges through a sphere of radius ' 𝟑𝒂′
with its centre at the origin?
(2013 All Delhi)
Charge Enclosed −2𝑄
Ans. Flux = 𝜀0
= 𝜀0

53. Two charges of magnitudes −𝟑𝑸 and +𝟐𝑸 are located at points (𝒂, 𝟎) and (𝟒𝒂, 𝟎)
respectively. What is the electric flux due to these charges through a sphere of radius ' 𝟓𝒂′
with its centre at the origin?
(2013 All India)

Charge Enclosed
Ans. Flux =
𝜀0
+2𝑄 − 3𝑄
=
𝜀0
−𝑄
=
𝜀0
54. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface
get affected when its radius is increased?
(2016 Delhi)
Ans. The electric flux due to a point charge enclosed by a spherical gaussian surface remains
'unaffected' when its radius is increased.

55. Does the charge given to a metallic sphere depend on whether it is hollow or solid? Give
reason for your answer.
(2017 Delhi)

Ans. No, it does not, because the charge resides only on the surface of the conductor.

56. A charge 𝒒 is enclosed by a spherical surface of radius 𝑹. If the radius is reduced to half,
how would the electric flux through the surface change?
(2009 All India)

Ans. As we know, Electric flux over an area in an electric field is the total number of lines of force
passing through the area. It is represented by 𝜙. It is a scalar quantity. Its S.I. unit is Nm2 C −1
or Vm.
⃗ ⋅ 𝑑S⃗ = 𝑞
i.e., 𝜙 = ∫s E 𝜀
0
Electric flux 𝜙 by 𝑞enclosed . So, it is invariant with radius R. Hence the electric flux through the
surface of sphere remains same.

57. A sphere 𝑺𝟏 of radius 𝒓_𝟏 encloses a net charge Q. if there is another concentric sphere 𝑺𝟐
of radius 𝒓𝟐 (𝒓𝟐 > 𝒓𝟏 ) enclosing charge 𝟐𝑸, find the ratio of the electric flux through 𝑺𝟏
and 𝑺𝟐 . How will the electric flux through sphere 𝑺𝟏 change if a medium of dielectric
constant 𝑲 is introduced in the space inside 𝐒𝟐 in place of air?

(2014 Comptt. All India)

 𝑄
Ans. Flux through 𝑆1 (𝜙1 ) = 𝜖 …..(i)
0

𝑄+2𝑄 3𝑄
Flux through 𝑆2 (𝜙2 ) = = ….(ii)
𝜖0 𝜖0
𝜙1 Q/𝜖0 1
∴ Ratio of flux = = =
𝜙2 3Q/𝜖0 3
Therefore, there will be no change in the flux through S1 on introducing dielectric medium
inside the sphere 𝑆2 .

58. Use Gauss's law to derive the expression for the electric field between two uniformly
charged large parallel sheets with surface charge densities 𝝈 and −𝝈 respectively.
(2009 All India)

Or

Two large parallel plane sheets have uniform charge densities +𝝈 and −𝝈. Determine the
electric field (i) between the sheets, and (ii) outside the sheets.
(2019 Series: BVM/4)

Ans. Gauss's Law. See Question 68.

𝜎
E = 2𝜀
0

In region I:
 𝜎
E1 = − 2𝜀 ,
0
𝜎
E2 = 2𝜀
0

Total field E1

= 𝐸1 + 𝐸2
−𝜎 𝜎
= 2𝜀 + 2𝜀 = 0
0 0

In region II:
𝜎 𝜎 𝜎
EII = 2𝜀 + 2𝜀 = 𝜀
0 0 0

In region III:
𝜎 𝜎
𝐸1 = 2𝜀 , 𝐸2 = − 2𝜀
0 0
𝜎 𝜎
∴ 𝐸 = 𝐸1 + 𝐸2 = + (− )=0
2𝜀0 2𝜀0

1.13 Applications of Gauss's Law

SA-I (1-2 Marks)

59. What is the direction of the electric field at the surface of a charged conductor having
charge density 𝝈 < 𝟎 ?
(2012 Comptt. Delhi)

Ans. The direction of electric field is normal and inward to the surface.

60. A spherical conducting shell of inner radius 𝒓𝟏 and outer radius 𝒓𝟐 has a charge ' 𝑸 '. A
charge ' 𝒒 ' is placed at the centre of the shell.
(a) What is the surface charge density on the (i) inner surface, (ii) outer surface of the
shell?
(b) Write the expression for the electric field at a point 𝒙 > 𝒓𝟐 from the centre of the shell.
(2010 All India)

Ans. (a) Surface charge density on the:

 −𝑞
(i) Inner surface of the shell, 𝜎in = 4𝜋𝑟2
1
Q+𝑞
(ii) Outer surface of the shell, 𝜎out =
4𝜋𝑟22

1 𝑞+Q
(b) Electric field at a point 𝑥 > 𝑟2 from the centre of the shell will be E = 4𝜋𝜀 ( 𝑥2
)
0

𝝈
61. ⃗⃗ = 𝒏ˆ,
Show that the electric field at the surface of a charged conductor is given by 𝑬 𝜺 𝟎
where 𝝈 is the surface charge density and 𝒏ˆ is a unit vector normal to the surface in the
outward direction.
(2010 All India)

1 Q
Ans. Electric field at a point on the surface of charged conductor, E = 4𝜋𝜀 2 .
0R
For simplicity we consider charged conductor as a sphere of radius ' 𝑅 '. If ' 𝜎 ' is in surface
charge density, then

1 4𝜋R2 𝜎 𝜎
Q = 4𝜋R2 𝜎 and E = 4𝜋𝜀 R2
=𝜀
0 0
𝜎
⃗
∴E = 𝜀 𝑛ˆ
0

...where [ 𝑛ˆ is a unit vector normal to the surface in the outward direction

62. A thin straight infinitely long conducting wire having charge density 𝝀 is enclosed by a
cylindrical surface of radius 𝒓 and length 𝒍, its axis coinciding with the length of the wire.
Find the expression for the electric flux through the surface of the cylinder.
(2011 All India)

𝐎𝐫
A thin straight infinitely long conducting wire of linear charge density ' 𝝀 ' is enclosed by a
cylindrical surface of radius ' 𝒓 ' and length ' 𝒍 '-its axis coinciding with the length of the
wire. Obtain the expression for the electric field, indicating its direction, at a point on the
surface of the cylinder. ( 2012 Comptt. Delhi)

Ans Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian
surface is zero. At the cylindrical part of the surface, E is normal to the surface at every
point, and its magnitude is constant, since it depends only on 𝑟. The surface area of the
curved part is 2𝜋𝑟𝑙, where 1 is the length of the cylinder.
Flux through the Gaussian surface
= Flux through the curved cylindrical part of the surface
= E × 2𝜋𝑟𝑙
 The surface includes charge equal to 𝜆𝑙.
Gauss's law then gives

E × 2𝜋𝑟𝑙 = 𝜆𝑙/𝜀0
𝜆
i.e., E=
2𝜋𝜀0𝑟

63 A hollow cylindrical box of length 𝟏 𝐦 and area of cross-section 𝟐𝟓 𝐜𝐦𝟐 is placed in a

three dimensional coordinate system as shown in the figure.

The electric field in the region is given by ⃗𝑬

⃗ = 𝟓𝟎𝒙𝒊ˆ where 𝐄 is in 𝐍𝐂−𝟏 and 𝒙 is in metres.
Find:
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.
(2013 Delhi)

Ans. (i) The magnitude of the electric field at the left face is E = 50NC−1
Therefore flux through this face

𝜙L = EAcos 𝜃
= 50 × 25 × 10−4 × cos 180∘
= −125 × 10−3 NC−1 m2
 The magnitude of the electric field at the right face is

E = 100NC −1

Therefore flux through this face

𝜙R = 100 × 25 × 10−4 × cos 0∘

= 250 × 10−3 NC−1 m2

Therefore net flux through cylinder is

𝜙𝑅 + 𝜙𝐿 = 125 × 10−3 NC −1 m2

Q
(ii) Charge enclosed by the cylinder 𝜙 = 𝜀
0

Q = 𝜙net × 𝜀0
= 125 × 10−3 × 8.856 × 10−12 C = 1107 × 10−15 C
𝐐 = 1.107𝑝C

64. Two large parallel plane sheets have uniform charge densities +𝝈 and −𝝈. Determine the
electric field (i) between the sheets, and (ii) outside the sheets.
(2019 Series: BVM/4)
+𝜎
Ans. Now electric field intensity due to a plane sheet of charge is given by 𝐸 = .
2𝜀0
+𝜎 −𝜎
Here E = 2𝜀 and E = 2𝜀
0 0
(i) Electric field at point 𝑄 (In between the sheets

⃗E = ⃗EA + 𝐸⃗B = 𝜎 + 𝜎 = 𝜎
2𝜀 02𝜀 𝜀
0 0
(ii) Field at the point 𝐏 or 𝐑 (outside the sheets)
𝜎 𝜎
⃗E = ⃗⃗⃗⃗
EA + ⃗⃗⃗⃗
EB = 2𝜀 − 2𝜀 = 0
0 0

65. Apply Gauss's law to show that for a charged spherical shell, the electric field outside the
shell is, as if the entire charge were concentrated at the centre.
(2019 Series: BVM/4)
Ans. Electric Field outside charged spherical shell. Flux through the small section of Gaussian
surface

𝜙=∮ →𝐸
⃗ ⋅ ⃗⃗⃗⃗
𝑑𝑠 ∴ 𝜙 = ∮ E ⋅ 𝑑𝑠cos 𝜃

∴ 𝐄 ∥ ⃗⃗⃗⃗
𝑑𝑠, 𝜃 = 𝟎 ∴ 𝜙 = E. 4𝜋R2 ….(i)
𝒒
Applying Gauss's theorem, 𝝓 = …….(ii)
𝜺𝟎
𝟏 𝒒
From equations (i) and (ii), 𝐄 = 𝟒𝝅𝜺 ⋅ 𝐑𝟐
𝟎
The expression indicates that the entire charge was concentrated at the centre.

SA-II
(3 Marks)

66. Draw a graph of electric field 𝐄(𝒓) with distance 𝒓 from the centre of the shell for 𝟎 ≤ 𝒓 ≤
∞.
(2009 Delhi)

Ans. Graph of electric filed E(r):

67. State Gauss' law in electrostatics. Using this law derive an expression for the electric field
due to a uniformly charged infinite plane sheet.
(2009 Delhi)
1
Ans. Gauss' Law states that "the total flux through a closed surface is 𝜀 times the net charge
0
enclosed by
𝑞
𝜙𝐸 = ∮ 𝐸⃗ ⋅ 𝑑𝑆 = ′
𝜀0

Let 𝜎 be the surface charge density (charge per unit area) of the given sheet and let P be a
point at distance 𝑟 from the sheet where we have to find ⃗E.

Choosing point P′ , symmetrical with P on the other side of the sheet, let us draw a Gaussian
cylindrical surface cutting through the sheet as shown in the diagram. As at the cylindrical
part of the Gaussian surface, E⃗ and 𝑑S⃗ are at a right angle, the only surfaces having E
⃗ and 𝑑S⃗
parallel are the plane ends

∴ 𝜙E = ∮ ⃗E ⋅ 𝑑S⃗ + ∮ ⃗E ⋅ 𝑑S⃗
= ∮ E𝑑 S + ∮ E𝑑 S = EA + EA = 𝟐𝐄𝐀
 ... [ As 𝐸⃗ is outgoing from both plane ends, the flux is positive. This is the total flux through
the Gaussian surface.
𝑞
Using Gauss' law, 𝜙E = 𝜀
0
𝑞 𝜎A
∴ 2EA = 𝜀 = 𝜀0
… [ As 𝒒 = 𝝈𝐀
0
𝜎
∴ 𝐸 = 2𝜀 .
0

This value is independent of 𝑟. Hence, the electric field intensity is same for all points near
the charged sheet. This is called uniform electric field intensity.

68. State 'Gauss law' in electrostatics. Use this law to derive an expression for the electric
field due to an infinitely long straight wire of linear charge density 𝝀𝐜𝐦−𝟏 .
(2009 Delhi, 2018, 2020 Series: HMJ/4)

Or
State Gauss' law. Using this law, obtain the expression for the electric field due to an
infinitely long straight conductor of linear charge density 𝝀. (2017 Comptt. Outside Delhi,
2023 Series: GEFH1/2)

Ans. Gauss's law in electrostatics. It states that "the total electric flux over the surface 𝑆 in
1
vaccum is times the total charge (q)."
𝜀0
𝑞
Contained in side S ∴ 𝜙E = ∮ s ⃗E ⋅ 𝑑S⃗ = 𝜀
0

Electric field due to an infinitely long straight wire: Consider an infinitely long straight line
charge having linear charge density 𝜆 to determine its electric field at distance 𝑟. Consider a
cylindrical Gaussian surface of radius 𝑟 and length 𝑙 coaxial with the charge. By symmetry,
the electric field E has same magnitude at each point of the curved surface 𝑆1 and is
directed radially outward.
Total flux through the cylindrical surface,

⃗ ⋅ 𝑑S⃗
∮ E =∮ ⃗ 𝑑S⃗⃗⃗1 + 𝑠∮ 𝑠 E
𝑠1 E ⋅
⃗ ⋅ 𝑑S⃗⃗⃗⃗2 + 𝑠∮ 𝑠 E⃗ ⋅ 𝑑S⃗⃗⃗⃗3
2 3
∘ ∘ ∘
=∮ 𝑠1 E𝑑𝑆1 cos 0 + E𝑑𝑆2 cos 90 + ∮ 𝑠3 E𝑑𝑆3 cos 90
𝑠2
= E∮ 𝑑𝑆1 = E × 2𝜋𝑟𝑙

As 𝜆 is the charge per unit length and is the length of the wire, so charge enclosed is,

𝑞 = 𝜆𝑙
 By Gauss's theorem,

⃗ ⋅ 𝑑S⃗⃗⃗1 = 𝑞 ⇒ E × 2𝜋𝑟𝑙 = 𝜆𝑙
∮ E 𝜀0 𝜀 0
𝜆
∴ E = 2𝜋𝜀
0𝑟

69. A positive point charge (+𝒒) is kept in the vicinity of an uncharged conducting plate.
Sketch electric field lines originating from the point on to the surface of the plate. Derive
the expression for the electric field at the surface of a charged conductor.
(2009 All India)
Ans . Representation of electric field. (due to a positive charge)

70 A charge is distributed uniformly over a ring of radius ' 𝒂 '. Obtain an expression for the
electric intensity 𝐄 at a point on the axis of the ring. Hence show that for points at large
distances from the ring, it behaves like a point charge.
(2016 Delhi)
Or
A thin circular ring of radius 𝒓 is charged uniformly so that its linear charge density
becomes 𝝀. Derive an expression for the electric field at a point 𝐏 at a distance 𝒙 from it
along the axis of the ring. Hence, prove that at large distance (𝒙 > 𝒓), the ring behaves as
a point charge.
(2020 Series: HMJ/5)

Ans. Electric Intensity on the axis of a ring:

 2𝜋𝑎
Net electric field at point 𝑃 = ∫0 𝑑𝐸cos 𝜃
Here, 𝑑E = Electric field due to a small element having charge 𝑑𝑞
1 𝑑𝑞
=
4𝜋𝜀0 𝑟2

𝑑𝑞
Let 𝜆 = Linear charge density = 𝑑𝑙
∴ 𝑑𝑞 = 𝜆𝑑𝑙
2𝜋𝑎 1 𝜆𝑑𝑙 𝑥 𝑥
Hence, E = ∫0 × … where [cos 𝜃 =
4𝜋𝜀0 𝑟2 𝑟 𝑟

𝜆𝑥
= (2𝜋𝑎)
4𝜋𝜀0𝑟 3
1 Q𝑥
= 3
4𝜋𝜀0 (𝑥 2
+𝑎 2)2

...where [ Total charge, 𝐐 = 𝝀 × 𝟐𝝅𝒂

At large distance, i.e., 𝑥 > 𝑎

1 Q
E = 4𝜋𝜀 ⋅ 𝑥 2
0

This is the Electric Field due to a point charge at distance 𝑥.

71. Two thin concentric and coplanar spherical shells, of radii 𝒂 and 𝒃(𝒃 > 𝒂) carry charges, 𝒒
and 𝑸, respectively. Find the magnitude of the electric field, at a point distant 𝒙, from
their common centre for
(i) 𝟎 < 𝒙 < 𝒂
(ii) 𝒂 ≤ 𝒙 < 𝒃
(iii) 𝒃 ≤ 𝒙 < ∞
(2016 Comptt. Delhi)

Ans. Magnitude of electric field. Two thin concentric and coplanar spherical shells of radii ' 𝑎 '
and ' 𝑏 ' (𝑏 > 𝑎) carry charges ' 𝑞 ' and ' 𝑄 ' respectively.

(i) For 0 < 𝑥 < 𝑎 Point lies inside both the spherical shells.
Hence, E(𝑥) = 0

(ii) For 𝑎 ≤ 𝑥 < 𝑏

Point is outside the spherical shell of radius ' 𝑎 ' but inside the spherical shell of radius ' 𝑏 '.

1 𝑞
∴ E(𝑥) = 4𝜋𝜀 ⋅ 𝑥 2
0
 (iii) For 𝑏 ≤ 𝑥 < ∞
Point is outside of both the spherical shells. Total effective charge at the centre equals (𝑄 +
𝑞).

1 (𝑞+Q)
∴ E(𝑥) = 4𝜋𝜀 ⋅ 𝑥2
0

72. State Gauss's law in electrostatics. Derive an expression for the electric field due to an
infinitely long straight uniformly charged wire.
(2020 Comptt Delhi)

1
Ans. Gauss Theorem. The surface integral of electric field over a closed surface is equal to 𝜀
0
times the charge enclosed by the surface.
𝑞
Alternatively, ∮ 𝐸⃗ ⋅ 𝑑𝑠 = 𝜀0

Expression for electric field

Flux through the Gaussian surface
= Flux through the curved
cylindrical part of the surface
= E × 2𝜋𝑟𝑙
Charge enclosed by the surface = 𝜆𝑙
𝜆𝑙
⇒ E × 2𝜋𝑟𝑙 =
𝜀0
𝜆
or E =
2𝜋𝜀0 𝑟

73 (i) An infinitely long positively charged straight wire has a linear charge density 𝝀. An
electron is revolving in a circle with a constant speed 𝒗 such that the wire passes through
the centre and is perpendicular to the plane, of the circle. Find the kinetic energy of the
electron in terms of magnitudes of its charge and linear charge density 𝝀 on the wire.
 (ii) Draw a graph of kinetic energy as a function of linear charge density 𝝀. (2023 Series:
GEFH1/2)
𝜆
Ans. (i) Electric Field, E =
2𝜋𝜀0 𝑟
Force, F = 𝑒E ….(i)
𝑚𝑣 2
and F = 𝑟
…..(ii)
From (i) and (ii), we get

𝑚𝑣 2
𝑒E = 2
𝑟 ⇒ 𝑒 𝜆 = 𝑚𝑣
𝑒𝜆 2𝜋𝜖0 𝑟 𝑟
⇒ 𝑚𝑣 2 =
2𝜋𝜖0
1
For K.E. multiply both sides by 2, we get

1 1 𝑒𝜆 𝒆𝝀
2
𝑚𝑣 2 = 2 2𝜋𝜖 ∴ K.E. = 𝟒𝝅𝝐
0 𝟎

(ii) Graph of Kinetic Energy:

𝑒𝜆
K=
4𝜋𝜀0
∴ K∝𝜆

74. A hollow conducting sphere of inner radius 𝒓𝟏 and outer radius 𝒓𝟐 has a charge 𝑸 on its
surface. A point charge +𝒒 is also placed at the centre of the sphere.
(a) What is the surface charge density on the (i) inner and (ii) outer surface of the sphere?
(b) Use Gauss' law of electrostatics to obtain the expression for the electric field at a point
lying outside the sphere.
(2020 Series: HMJ/4)

Ans. (a) Surface charge density on the:

(i) inner surface of the shell,
−𝑞
𝜎in = 4𝜋𝑟2
1
 (ii) outer surface of the shell,

Q+𝑞
𝜎out =
4𝜋𝑟22

(b) Derivation of expression of electric field for a spherical Gaussian surface when 𝑥 > 𝑟2 .
As per Gauss' law,

Flux (𝜙) ⃗ 𝑑𝑆 = (Q+𝑞)

=∮ E 𝜀 0
2 (Q+𝑞) 1 (Q+𝑞)
⇒ E × 4𝜋𝑥 = 𝜀0
∴ E = 4𝜋𝜀 𝑥2
0

75. An infinitely large thin plane sheet has a uniform surface charge density +𝝈. Obtain the
expression for the amount of work done in bringing a point charge 𝒒 from infinity to a
point, distant 𝒓, in front of the charged plane sheet.
(2017 Outside Delhi)
Ans.
𝑟
W = 𝑞 ∫∞ 𝐸⃗ ⋅ 𝑑𝑟
𝑟 𝑟 𝜎 𝑞𝜎
= 𝑞 ∫∞ (−𝐸𝑑𝑟) = 𝑞 ∫∞ ( ) 𝑑𝑟 = [∞ − 𝑟]
2𝜀0 2𝜀
⇒ (∞)

76. A wire 𝑨𝑩 of length 𝑳 has linear charge density 𝝀 = 𝒌𝒙, where 𝒙 is measured from the end
𝑨 of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for
the electric flux through this surface. (2017 Comptt. Outside Delhi)
Ans. Given: Length of wire = L, Charge density (𝜆) = 𝑘𝑥, 𝜙 = ?
We know,

𝑑𝑞 = 𝜆𝑑𝑥 = 𝑘𝑥𝑑𝑥
𝑞 L 1
Q = ∫0 𝑑𝑞 = ∫0 𝑘𝑥𝑑𝑥 = 2 𝑘 L2
Q 𝑘 L2
∴ 𝜙 =𝜖 = 2𝜖0
0

77. (a) A small metal sphere carrying charge +𝑸 is located at the centre of a spherical cavity
inside a large uncharged metallic spherical shell as shown in the figure. Use Gauss's law to
find the expressions for the electric filed at points 𝑷𝟏 and 𝑷𝟐.
 (b) Draw the pattern of electric field lines in this arrangement.
(2012 Comptt. All India)

Ans. (a) Calculation of electric field at point 𝑃1 :

Net charge enclosed by the Gaussian surface is +Q.

Q
∴ 𝜙 = ∮ 𝐸⃗ ⋅ 𝑑𝑠 = 𝜀
0

As electric field of positive charge is radially outwards, it is parallel to the area vector on the
surface chosen.

Q
∴ ∮ ⃗E ⋅ 𝑑𝑠 = ∮ E ⋅ 𝑑𝑠cos 0∘ = 𝜀
0
Q 2 Q
E∮ 𝑑𝑠 = 𝜀 ⇒ E × 4𝜋𝑟 = 𝜀
0 0
Q
∴ E = 4𝜋𝜀 2
0𝑟

As point P2 lies inside the metal, therefore electric field at point 𝑃2 is zero.

(b)
78. Using Gauss' law deduce the expression for the electric field due to a uniformly charged
spherical conducting shell of radius 𝐑 at a point ( 𝒊 ) outside and (ii) inside the shell. Plot a
graph showing variation of electric field as a function of 𝒓 > 𝐑 and 𝒓 < 𝐑 ( 𝒓 being the
distance from the centre of the shell).
(2013 All India)
Or
A thin conducting spherical shell of radius 𝑹 has charge 𝐐 spread uniformly over its
surface. Using Gauss's law, derive an expression for an electric field at a point outside the
shell.
(2009 Delhi)
Ans. (i) Field Outside Shell: Consider a thin spherical shell of radius R with centre O. Let charge
+𝑞 be distributed uniformly over the surface of shell. To calculate electric field intensity at P
where OP = 𝑟, imagine a sphere S, with centre at O and radius 𝑟.

The surface of sphere is Gaussian surface over at every point. Electric field is same and directed
radially outwards.
Applying Gauss' theorem,

⃗⃗⃗⃗ = 𝑞 ⇒ E =
∫S 𝐸⃗ ⋅ 𝑑𝑆
𝑞
2
𝜀
0 𝜀 0 4𝜋𝑟
𝑞
⇒ E= … [𝐒 = 𝟒𝝅𝒓𝟐
4𝜋𝜀0𝑟 2

[𝑟 is distance of point P from centre where E is calculated.]

(ii) Inside Shell. As we know charge is located on its surface,
 𝑞
∴ E. 𝑑𝑆 = 𝜀 (𝑞 = 0)
0

Hence, 𝐸⃗ = 0

1
(iii) at 𝑟 < R, ⃗E is zero and at 𝑟 = R, E is maximum at 𝑟 > R, E is decreasing at E ∝ 𝑟2

79. (a) Use Gauss's law, prove that the electric field at a point due to a field at a point due to a
uniformly charged infinite plane sheet in independent of the distance from it.
(b) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?
(2012 Delhi)
Ans. (a) Consider a thin, infinite plane sheet of charge with uniform surface charge density 𝜎. We
wish to calculate its electric field at a point P at distance 𝑟 from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same
magnitude and opposite direction at two points P anel P′ equidistant from the sheet and on
opposite sides. We choose cylindrical Gaussian surface of cross-sectional area A and length
2𝑟 with its axis perpendicular to the sheet.
As the lines of force are parallel to the curved surface of the cylinder, the flux through the
curved surface is zero. The flux through the plane-end faces of the cylinder is:

𝜙E = EA + EA = 2EA
 Charge enclosed by the Gaussian surface,

𝑞 = 𝜎A

According to Gauss's theorem,

𝑞 𝜎A 𝜎A
𝜙E = 𝜀 ∴ 2EA = 𝜀0
or E = 2𝜀
0 0

Clearly, E is independent of 𝑟, the distance from the plane sheet.

(b)

(i) For positively charged sheet → away from the sheet

(ii) For negatively charged sheet → towards the sheet

LA (5 Marks)

80. (a) Using Gauss' law, derive an expression for the electric field intensity at any point
outside a uniformly charged thin spherical shell of radius 𝑹 and charge density 𝝈𝐂/𝐦𝟐.
Draw the field lines when the charge density of the sphere is (i) positive, (ii) negative.
(b) A uniformly charged conducting sphere of 𝟐. 𝟓 𝐦 in diameter has a surface charge
density of 𝟏𝟎𝟎𝝁𝐂/𝐦𝟐. Calculate the
(i) charge on the sphere.
(ii) total electric flux passing through the sphere.
(2008 Delhi)

Ans. (a) (i) To find out electric field at a point outside a spherical charged shell we imagine a
symmetrical Gaussian surface in such a way that the point lies on it.
𝑞
From Gauss's theorem, 𝜙 = ∮ S 𝐸⃗ ⋅ 𝑑𝑆 = 𝜀𝑚
0
Flux 𝜙 through 𝑆 ′

𝜙=∮ ⃗
S′ 𝐸 ⋅ 𝑑𝑆 =∮ ⃗ ⋅ 𝑑𝑆 = E. 4𝜋𝑟 2
𝑆′ 𝐸
𝑞
⇒ E⋅ 4𝜋𝑟 2 = 𝜀𝑚
0
1 𝑞𝑚
⇒ E= ⋅
4𝜋𝜀0 𝑟2
(ii)

2.5
(b) (i) Given: 𝑟 = m, 𝜎 = 100𝜇C/m2
2
Charge on the sphere, Q = 𝜎. 4𝜋𝑟 2
2.5 2
or Q = 100 × 10−6 × 4 × 3.14 × ( 2 )
= 19.6 × 10−4 C = 1.96 × 10−3 C

(ii) Flux passing through the sphere

Q 19.6×10−4
𝜙 = 𝜀 or 𝜙 = 8.85×10−12
0
∴𝜙 = 2.2 × 108 Nm2 /C

81. Using Gauss's law, derive the expression for the electric field at a point (i) outside and (ii)
inside a uniformly charged thin spherical shell. Draw a graph showing electric field 𝐄 as a
function of distance from the centre.
(2013 Comptt. All India)

Ans. Electric field due to a uniformly charged spherical shell:

Suppose a thin spherical shell of radius R and centre O.
Let the charge +𝑞 be distributed over the surface of sphere.
Electric field intensity 𝐸⃗ is same at every point on the surface of sphere directed directly
outwards.
 Let a point P be outside the shell with radius vector 𝑟 and small area element ⃗⃗⃗⃗
𝑑𝑆 = 𝑛ˆ 𝑑𝑆
According to the Gauss's law

⃗⃗⃗⃗ = 𝑞
⃗ ⋅ 𝑑𝑆
∮ E 𝜀 0
𝑞
⇒ ∮ E𝑑 S =
𝜀0

Since ⃗E and 𝑛ˆ are in the same direction

𝑞
∴ E4𝜋𝑟 2 = 𝜀
0
1 𝑞 𝟏 𝒒
⇒ E= ⃗ =
… where [ Vectorially 𝐄 𝒓
4𝜋𝜀0 𝑟2 𝟒𝝅𝜺𝟎 𝒓𝟐

(i) If 𝜎 is the surface charge density on the shell, then 𝑞 = 4𝜋R2

1 4𝜋R2 𝜎 𝜎
∴ E = 4𝜋𝜀 R2
=𝜀
0 0

(ii) If the point P lies inside the spherical shell, then the Gaussian surface encloses no charge.

∴𝑞=0

Hence E = 0

82. (a) "Gauss's law in electrostatics is true for any closed surface, no matter what its shape or
size is." Justify this statement with the help of a suitable example.

(b) Use Gauss's law to prove that the electric field inside a uniformly charged spherical shell is
zero.
(2015 All India)

Ans. (a) Gauss's Law states that the electric flux through a closed surface is given by
𝑞
𝜙=𝜖
0
The law implies that the total electric flux through a closed surface depends on the quantity of total
charge enclosed by the surface, and does not depend on its shape and size. For example, net charge
enclosed by the electric dipole (𝑞, −𝑞) is zero, hence the total electric flux enclosed by a surface
containing electric dipole is zero.

(b) Electrical field inside a uniformly charged spherical shell. Let us consider a point ' P ' inside the
shell. The Gaussian surface is a sphere through P centred at O.
The flux through the Gaussian surface is

E × 4𝜋𝑟 2 .

However, in this case, the Gaussian surface encloses no charge.

Gauss's law then gives

E × 4𝜋𝑟 2 = 0
or E=0
(𝑟 < R)

that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.

2024 CBSE Question Paper (RQSP4/4)

Set-1: 55/4/1

Q.2. Two identical small conducting balls 𝑩𝟏 and 𝑩𝟐 are given −𝟕𝐩𝐂 and +𝟒𝐩𝐂 charges
respectively. They are brought in contact with a third identical ball 𝑩𝟑 and then separated.
If the final charge on each ball is −𝟐𝐩𝐂, the initial charge on 𝐁𝟑 was
1
(a) −2pC
(b) −3pC
(c) −5pC
(d) −15pC

Q.32. Or, (b) (i) Using Gauss's law, show that the electric field 𝐸⃗ at a point due to a uniformly
𝜎
charged infinite plane sheet is given by ⃗E = 𝑛ˆ where symbols have their usual meanings.
2𝜀0

(ii) Electric field ⃗E in a region is given by ⃗E = (5𝑥 2 + 2) 𝑖ˆ where 𝐸⃗ is in N/C and 𝑥 is in

meters.
 A cube of side 10 cm is placed in the region as shown in figure.
Calculate (1) the electric flux through the cube, and (2) the net charge enclosed by the cube.

Set-II : 55/4/2

1
Q.2. The Coulomb force (F) versus (𝑟2 ) graphs for two pairs of point charges ( 𝑞1 and 𝑞2 ) and (
𝑞2 and 𝑞3 ) are shown in figure. The charge 𝑞2 is positive and has least magnitude. Then

(a) 𝑞1 > 𝑞2 > 𝑞3

(b) 𝑞1 > 𝑞3 > 𝑞2
(c) 𝑞3 > 𝑞2 > 𝑞1
(d) 𝑞3 > 𝑞1 > 𝑞2

Set-III : 55/4/3

Q.1. An electric dipole of dipole moment p⃗ is kept in a uniform electric field 𝐸⃗. The amount of
work done to rotate it from the position of stable equilibrium to that of unstable equilibrium
will be.
(a) 2pE
(b) −2pE
 (c) pE
(d) zero

Q.2. An infinite long straight wire having a charge density 𝜆 is kept along 𝑦 axis in 𝑥𝑦 plane. The
Coulomb force on a point charge 𝑞 at a point P(𝑥, 0) will be

𝑞𝜆 𝑞𝜆
(a) attractive and 2𝜋𝜀 (b) repulsive and 2𝜋𝜀
0𝑥 0𝑥
𝑞𝜆 𝑞𝜆
(c) attractive and (d) repulsive and
𝜋𝜀0 𝑥 𝜋𝜀0𝑥

Q.23. Two conducting spherical shells A and B of radii R and 2R are kept far apart and charged to
the same charge density 𝜎. They are connected by a wire. Obtain an expression for final
potential of shell A. 3

2024 CBSE Question Paper (SRQP/5)

Set-I: 55/5/1

Q.17. (a) Four point charges of 1𝜇C, −2𝜇C, 1𝜇C and −2𝜇C are placed at the corners A, B, C and D
respectively, of a square of side 30 cm. Find the net force acting on a charge of 4𝜇C placed
at the centre of the square.

Or

(b) Three point charges, 1pC each, are kept at the vertices of an equilateral triangle of side
10 cm. Find the net electric field at the centroid of triangle.

Q.23. (a) Define the term 'electric flux' and write its dimensions.
(b) A plane surface, in shape of a square of side 1 cm is placed in an electric field ⃗E =
N
(100 ) 𝑖ˆ such that the unit vector normal to the surface is given by 𝑛ˆ = 0.8𝑖ˆ + 0.6𝑘ˆ .
C
Find the electric flux through the surface. 3

Q.31. (b) (i) A thin spherical shell of radius R has a uniform surface charge density sigma ( 𝜎 ) Using
Gauss' law, deduce an expression for electric field (i) outside and (ii) inside the shell.
(ii) Two long straight thin wires AB and CD have linear charge densities 10𝜇C/m and
−20𝜇C/m, respectively. They are kept parallel to each other at a distance 1 m.

Find magnitude and direction of the net electric field at a point midway between them.
5

Set-II : 𝟓𝟓/𝟓/𝟐

Q.13. Assertion (A): Equal amount of positive and negative charges are distributed uniformly on
two halves of a thin circular ring as shown in figure. The resultant electric field at the centre
𝑂 of the ring is along OC. 1
 Reason (R): It is so because the net potential at 𝑂 is not zero.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of
Assertion (A).
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation
of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Both Assertion (A) and Reason (R) are false.

Competency Based Questions

MCQ Multiple Choice Questions

1 SI unit of permittivity of free space is

(a) Farad
(c) C2 N −1 m−2
(b) Weber
(d) C2 N −1 m−2
2 A charge 𝑸 is placed at the centre of the line joining two point charges +𝒒 and +𝒒 as
shown in the figure. The ratio of charges 𝑸 and 𝒒 is

(a) 4
(b) 1/4
(c) -4
(d) −1/4
3 The force per unit charge is known as
(a) electric flux
(b) electric current
(c) electric potential
(d) electric field
4 Electric field lines provide information about
(a) field strength
(b) direction
(c) nature of charge
(d) all of these
5 Which of the following figures represent the electric field lines due to a single negative
charge?

6 The SI unit of electric flux is

(a) NC −1 m−2
(b) NCm−2
(c) NC−2 m2
(d) NC −1 m2
7 The unit of electric dipole moment is
(a) newton
(b) coulomb
(c) farad
(d) debye
8 Consider a region inside which, there are various types of charges but the total charge is
zero. At points outside the region
(a) the electric field is necessarily zero.
(b) the electric field is due to the dipole moment of the charge distribution only.
(c) the dominant electric field is inversely proportional to 𝑟 3 , for large 𝑟 (distance from
origin).
(d) the work done to move a charged particle along a closed path, away from the region will
not be zero.
9 The surface considered for Gauss's law is called
(a) Closed surface
(b) Spherical surface
(c) Gaussian surface
(d) Plane surface
10 The total flux through the faces of the cube with side of length 𝒂 if a charge 𝒒 is placed at
corner A of the cube is
 𝑞 𝑞
(a) 8𝜀 (b) 4𝜀
0 0
𝑞 𝑞
(c) 2𝜀 (d) 𝜀
0 0

11. Which of the following statements is not true about Gauss's law?
(a) Gauss's law is true for any closed surface.
(b) The term 𝑞 on the right side of Gauss's law includes the sum of all charges enclosed by
the surface.
(c) Gauss's law is not much useful in calculating electrostatic field when the system has some
symmetry.
(d) Gauss's law is based on the inverse square dependence on distance contained in the
coulomb's law.

12. An isolated point charge particle produces an electric field ⃗𝑬

⃗ at a point 𝟑 𝐦 away from it.
⃗
𝑬
The distance of the point at which the field is 𝟒 will be
(a) 2 m (b) 3 m
(c) 4 m (d) 6 m
(2023 GEFH1/2)

13. A steady current of 𝟖 𝐦𝐀 flows through a wire. The number of electrons passing through
a cross-section of the wire in 𝟏𝟎 𝐬 is
(2023 GEFH1/2)
(a) 4.0 × 1016 (b) 5.0 × 1017
(c) 1.6 × 1016 (d) 1.0 × 1017

14. A point charge 𝒒𝟎 is moving along a circular path of radius a, with a point charge −𝑸 at the
centre of the circle. The kinetic energy of 𝒒𝟎 is
(2023 GEFH1/2)
𝑞0 Q 𝑞0 Q
(a) (b)
4𝜋𝜖0 𝑎 8𝜋𝜖0 𝑎
𝑞0 Q 𝑞0 Q
(c) (d)
4𝜋𝜖0𝑎 2 8𝜋𝜖0 𝑎 2
15 The magnitude of the electric field due to a point charge object at a distance of 𝟒. 𝟎 𝐦 is
𝟗 𝐍/𝐂. From the same charged object the electric field of magnitude, 𝟏𝟔 𝐍/𝐂 will be at a
distance of
(2023 GEFH1/2)
(a) 1 m (b) 2 m
(c) 3 m (d) 6 m

16 A point 𝐏 lies at a distance 𝒙 from the mid point of an electric dipole on its axis. The
electric potential at point 𝑷 is proportional to
(2023 GEFH1/2)
1 1
(a) 𝑥 2 (b) 𝑥 3
1 1
(c) (d)
𝑥4 𝑥 1/2

17 An electric dipole of length 𝟐 𝐜𝐦 is placed at an angle of 𝟑𝟎∘ with an electric field 𝟐 ×

𝟏𝟎𝟓 𝐍/𝐂. If the dipole experiences a torque of 𝟖 × 𝟏𝟎−𝟑 𝐍𝐦, the magnitude of either
charge of the dipole, is
(2023 GEFH1/5)
(a) 4𝜇C (b) 7𝜇C
(c) 8mc (d) 2mC
18 A charge 𝑸 is placed at the centre of a cube. The electric flux through one if its face is
(2023 GEFH1/5)
Q 𝑄
(a) 𝜀 (b) 6𝜀
0 0
𝑄 Q
(c) (d)
8𝜀0 3𝜀0

19 A negatively charged object 𝑿 is repelled by another charged object Y. However an object

𝐙 is attracted to object 𝐘. Which of the following is the most possibility for the object 𝒁 ?
(2021 Term 1)
(a) positively charged only
(b) negatively charged only
(c) neutral or positively charged
(d) neutral or negatively charged
20 In an experiment three microscopic latex spheres are sprayed into a chamber and became
charged with charges +𝟑𝒆, +𝟓𝒆 and −𝟑𝒆 respectively. All the three spheres came in
contact simultaneously for a moment and got separated. Which one of the following are
possible values for the final charge on the spheres?
(2021 Term 1)
(a) +5𝑒, −4𝑒, +5𝑒 (b) +6𝑒, +6𝑒, −7𝑒
(c) +4𝑒, +3.5𝑒, +5.5𝑒 (d) +5𝑒, −8𝑒, +7𝑒
21 An object has charge of 𝟏𝐂 and gains 𝟓. 𝟎 × 𝟏𝟎𝟏𝟖 electrons. The net charge on the object
becomes:
(2021 Term 1)
(a) −0.80C (b) +0.80C
(c) +1.80C (d) +0.20C
 22 Which of the diagrams correctly represents the electric field between two charged plates if
a neutral conductor is placed in between the plates?
(2021 Term 1)

23 The magnitude of electric field due to a point charge 𝟐𝒒, at distance 𝒓 is 𝑬. Then the
magnitude of electric field due to a uniformly charged thin spherical shell of radius 𝑹 with
𝒓
total charge 𝒒 at a distance 𝟐 ( 𝒓 > 𝐑 ) will be
(2021 Term 1)
𝐸
(a) (b) 0
4
(c) 2E (d) 4E
24 A square sheet of side ' 𝒂 ' is lying parallel to 𝐗𝐘 plane at 𝒛 = 𝒂. The electric field in the
⃗ = 𝒄𝒛𝟐 𝒌ˆ. The electric flux through the sheet is
region is ⃗𝑬
(2021 Term 1)
1
(a) 𝑎4 𝑐 (b) 3 𝑎3 𝑐
1
(c) 𝑎4 𝑐 (d) 0
3

25 Three charges 𝒒, −𝒒 and 𝒒𝟎 areplaced as shown in figure. The magnitude of the net force
𝟏
on the charge 𝒒𝟎 at point 𝑶 is [𝒌 = (𝟒𝝅𝜺 )].
𝟎
(2021 Term 1)

2𝑘𝑞𝑞0
(a) 0 (b) 𝑎2
√2𝑘𝑞𝑞0 1 𝑘𝑞𝑞0
(c) 𝑎2
(d)
√2 𝑎 2

26. Four objects 𝑾, 𝑿, 𝒀 and 𝒁, each with charge +𝒒 are held fixed at four points of a square
of side 𝒅 as shown in the figure. Objects 𝑿 and 𝒁 are on the midpoints of the sides of the
 square. The electrostatic force exerted by object 𝑾 on object 𝑿 is 𝑭. Then the magnitude
of the force exerted by object 𝑾 on 𝒁 is
F F
(a) 7 (b) 5
F F
(c) 3 (d) 2

ARQ Assertion-Reason Questions

DIRECTION: Read the two statements Assertion (A) and Reason (R) carefully to mark the
correct option out of the options given below:
(a) Assertion and Reason both are correct statements and Reason is correct explanation for
Assertion.
(b) Assertion and Reason both are correct statements but Reason is not correct explanation
for Assertion.
(c) Assertion is correct statement but Reason is wrong statement.
(d) Assertion is wrong statement but Reason is correct statement.

1. Assertion: If there exists coulomb attraction between two bodies, both of them may not be
charged.
Reason: In coulomb attraction two bodies are oppositely charged.
2 Assertion: Electric force acting on a proton and an electron, moving in a uniform electric
field is same, where as acceleration of electron is 1836 times that a proton.
Reason: Electron is lighter than proton.
3 Assertion: As force is a vector quantity, hence electric field intensity is also a vector quantity.
Reason: The unit of electric field intensity is newton per coulomb.
4 Assertion: The surface densities of two spherical conductors of different radii are equal.
Then the electric field intensities near their surface are also equal.
Reason: Surface charge density is equal to charge per unit area.
5 Assertion: The electric lines of forces diverges from a positive charge and converge at a
negative charge.
Reason: A charged particle free to move in an electric field always move along an electric
line of force.
6 Assertion: Range of Coulomb force is infinite.
Reason: Coulomb force acts between two charged particles.
7 Assertion: A small metal ball is suspended in a uniform electric field with an insulated
thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the electric
field.'
Reason: X-rays emits photo electron and metal becomes negatively charged.
8 Assertion: If a point charge be rotated in a circle around a charge, the work done will be
zero.
Reason: Work done is equal to dot product of force distance.
9 Assertion: No two electric lines of force can intersect each other.
Reason: Tangent at any point of electric line of force gives the direction of electric field.
 10. Assertion: Sharper is the curvature of spot on a charged body lesser will be the surface
charge density at that point.
Reason: Electric field is non-zero inside a charge conductor.

11. Assertion: A negative charge in an electric field moves along the direction of the electric
field.
Reason: On a negative charge a force acts in the direction of the electric held.
(2021 Term 1)

CBQ
Case-Based Questions
I. Read the para given below and answer the questions that follow:
A charge is a property associated with the matter due to which it experiences and produces an
electric and magnetic field. Charges are scalar in nature and they add up like real number. Also, the
total charge of an isolated system is always conserved. When the objects rub against each other
charges acquired by them must be equal and opposite.

1. The cause of a charing is:

(a) The actual transfer of protons.
(b) The actual transfer of electrons.
(c) The actual transfer of neutrons.
(d) None the above

2. Pick the correct statement.

(a) The glass rod gives protons to silk when they are rubbed against each other.
(b) The glass rod gives electrons to silk when they are rubbed against each other.
(c) The glass rod gains protons from silk when they are rubbed against each other.
(d) The glass rod gains electrons when they are rubbed against each other.

3. If two electrons are each 𝟏. 𝟓 × 𝟏𝟎−𝟏𝟎 𝐦 from a proton, the magnitude of the net electric force
they will exert on the proton is
(a) 1.97 × 10−8 N (b) 2.73 × 10−8 N
−8
(c) 3.83 × 10 N (d) 4.63 × 10−8 N
4. A charge is a property associated with the matter due to which it produces and experiences:
(a) electric effects only
(b) magnetic effects only
(c) both electric and magnetic effects
(d) none of these.

5. The cause of quantization of electric charges is:

(a) Transfer of an integral number of neutrons.
(b) Transfer of an integral number of protons.
(c) Transfer of an integral number of electrons.
(d) None of the above.
II. Read the para given below and answer the questions that follow:
Surface Charge Density. Surface charge density is defined as the charge per unit surface area the
surface (Arial) charge symmetric distribution and follow Gauss law of electrostatics mathematical
term of surface charge density σ=ΔQ/ΔS.

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have
surface charge densities of opposite sign ( ±𝜎 ). Having magnitude 8.8 × 10−12 cm−2 as shown
𝜎
⃗ ΔS⃗, where Δ𝑆 = 1 m2 (unit
here. The intensity of electrified at a point is 𝐸 = and flux is 𝜙 = E
𝜀0
arial plate).

6. 𝑬 in the outer region (I) of the first (A) plate is

(a) 1.7 × 10−22 N/C (b) 1.1 × 10−12 V/m
(c) Zero (d) Insufficient data

7. E in the outer region (III) of the second plate (B) is

(a) 1 N/C (b) 0.1 V/m
(c) 0.5 N/C (d) zero

8. E between (II) the plate is

(a) 1 N/C (b) 0.1 V/m
(c) 0.5 N/C (d) None of these
9. The ratio of 𝑬 from left side of plate 𝑨 at distance 𝟏 𝐜𝐦 and √𝟐 𝐦 respectively is
(a) 1: √2 (b) 10: √2
(c) 1: 1 (d) √20: 1

10. In order to estimate the electric field due to a thin finite plane metal plate the Gaussian
surface considered is
(a) Spherical (b) Linear
(c) Cylindrical (d) Cubic

III. Read the para given below and answer the questions that follow:
1
Gauss Theorem. The total flux through a closed surface, enclosing a volume, in vacuum is 𝜀 time the
0
net change, enclosed by the surface.
𝑞
𝜙 = ∮ 𝐬 ⃗E × 𝑑S⃗ = enclosed
𝜀 0

Gaussian Surface. Any closed surface imagined around the charge distribution, so that Gauss
theorem can be conveniently applied to find electric field due to the give charge distribution.
Electric field due to infinitely long straight charged wire of linear charge density 𝜆;
𝜆
E = 2𝜋𝜀 𝑟 , where 𝑟 is the perpendicular distance of the observation point from the wire. Electric
0
field due to an infinite plane sheet of charge of surface charge density 𝜎.
𝜎
E = 2𝜀
0

11. S.I unit of electric flux is …………. .

(a) N2 m, C (b) NmC−2
(c) Nm2 C−1 (d) Nm−2 C

12. Electric flux is a …………. .

(a) Constant quantity (b) Vector quantity
(c) Scalar quantity (d) None of these

13. Two charges of magnitude −𝟐𝑸 and +𝑸 are located at points (𝒂, 𝟎) and (𝟒𝒂, 𝟎) respectively.
What is the electric flux due to these charges through a sphere of radius ' 𝟑𝒂′ ' with its centre at
origin?
𝑄 −2𝑄
(a) (b)
𝜀0 𝜀0
3𝑄 −3𝑄
(c) 𝜀0
(d) 𝜀0

14. A charge 𝒒 is placed at the centre of a cube of side 𝒍. What is the electric flux passing through
each face to the cube?
𝑞 𝑞
(a) (b)
5𝜀0 9𝜀0
𝑞 𝑞
(c) 6𝜀 (d) 𝜀
0 0
15. Three charges +𝟐𝒒, −𝒒 and +𝟑𝒒 are given. Two charges +𝟐𝒒 and −𝒒 are enclosed with in a
surface ' 𝐒 '. What is the electric flux due to this configuration through the surface ' 𝐒 ' ?
5𝑞 3𝑞
(a) 2𝜀 (b) 𝜀
0 0
4𝑞 𝑞
(c) 𝜀0
(d) 𝜀
0