Solution

TEST 02 (BOARD LEVEL)

Class 12 - Physics
Section A
1.
(b) decrease
Explanation:
Since the spheres are conducting, the surface charge distribution on each sphere will be
altered because of the repulsion from the charges on the other sphere. In particular, the
charges on each sphere will be pushed away by the charges on the other sphere. This
will cause the charges on opposite spheres to be further away from each other, and the
force of repulsion to be less than in the case of a uniform surface charge distribution.
2.
(b) decreases K times
Explanation:
when air is replaced by dielectric medium, electrostatic force decreases by K times
3.
(d)
−Q

Explanation:

The total force on one Q is

2
kQ kqQ
F = +
2
4a a2

For the system to be in equilibrium F = 0

2
kQ kqQ
+ = 0
2
4a a2

Q
q = −
4

4. (a)
ρr

3ε0

Explanation:
Electric field inside a uniformly charged sphere (r < R),
E= 1 q
⋅ 3
r
4πε0 R

But q = 4

3
π R3ρ
E=
ρr
∴
3ε0

5.
(c) λ

2πε0 a

Explanation:
λ= linear charge density;
Charge on elementary portions is given by dq = λdx
 Electric field at O is given by , dE =
λdx

4πε0 a
2

Horizontal electric field, i.e., perpendicular to AO, will cancelled.

Hence, net electric field = addition of all electrical fields in direction of AO
= ∑ dE cos θ

λdx
⇒ E = ∫ cos θ
2
4πε0 a

Also, dθ = dx

a
or dx = adθ
π
π/2 λ cos θdθ λ 2
E = ∫ = [sin θ]
−π/2 4πε0 a 4πε0 a −π/2

λ λ
= [1 − (−1)] =
4πε0 a 2πε0 a

6.
(d) electric field intensity
Explanation:
The relation between E, σ and E is E = σ

7. (a) 3 NC-1
Explanation:
−11
-1
E= σ

ε0
=
2.56×10

8.854×10
−12
≃ 3 NC
8.
(c) A is true but R is false.
Explanation:
A is true but R is false.
9.
(b) Assertion and reason both are correct statements but reason is not correct
explanation for assertion.
Explanation:
Coulomb attraction exists even when one body is charged and the other is uncharged.
Section B
10. Upward electric force on particle = Weight of the particle
mg = qE = q ⋅ 2ε0
σ

or q =
2ε0 mg

−12 −9
2×8.85×10 ×5×10 ×9.8
=
−6
4×10

-13
= 2.16 × 10 C
Number of electrons required to be removed,
6 −13

n= = = 1.355 × 10
q 2.16×10

e −19
1.6×10

4 2
11. Here q = 2.0 × 10-4 C, m = 10g = 10-2 kg
r = 10 cm = 0.10 m
Let v be the speed of each particle at infinite separation. By conservation of energy,
P.E. of two particles at the separation of 10 cm = K.E. of the two particles at infinite
separation
1 q q 1 1
1 2 2 2
⋅ = mv + mv
4πε0 r 2 2

or v 1 q1 ,q2
2
= ,
4πε0 rm

9 −4 −4
4
=
9×10 ×2.0×10 ×2.0×10
−2
= 36 × 10
0.10×10

v = 600 ms-1
∴

12. In fig. two charges are -q are held at a distance of 2d. Horizontal component of the forces
cancel each other and the vertical component gives the net amount of force. So , Net
force F on q towards the centre O
2 2
q 2q x
F = 2 2
cos θ = − 2
⋅
4πεo r 4πε0 r r

2
2q x
F = − 3/2
4πεo 2 2
(d +x )

2
2q
≈ − 3
x = −k; for x << d
4πεo d

Thus, the force on the third charge q is proportional to the displacement and is towards
the centre of the two other charges.
Therefore, the motion of the third charge is harmonic with frequency
−−
¯
K
ω = √
m

−−
2π m
⇒ T = = 2π√
ω K

−−−−−−
3 3 3 1/2
m.4πε0 d 8π ε0 md
⇒ T = 2π√ 2
= [ 2
]
2q q

13.

Now Electric field Intensity due to a plane sheet of charge

σ
E =
2ε0

Here E A
=
+σ

2ε0
and E B
=
−σ

2ε0

i. Electric field at Point Q (In between the sheets)

⃗ ⃗ ⃗ σ σ σ
E = EA + EB = + =
2ε0 2ε0 ε0

ii. Field at the point P or R(outside the sheet).

⃗ ⃗ ⃗ σ σ
E = EA + EB = − = 0
2ε0 2ε0

14. Here, q = 1.6 × 10-7 C; the radius of the sphere; R = 12 cm = 0·12 cm

i. The charge on a conductor resides on its outer surface. Therefore, the electric field
inside the sphere is zero.
 ii. For a point on the charged spherical conductor or outside it, the charge may be
assumed to be concentrated at its centre. Therefore, electric field just outside the
sphere
= 105 NC-1
−7

= 1 q 9 1⋅6×10
⋅ 2
= 9 × 10 × 2
4πε0 R (0.12)

iii. Here, r = 1.8 cm = 0.18 m

Electric field at a point 0.18 m from the centre
q −7
1 9 1⋅6×10
= ⋅ = 9 × 10 ×
4πε0 r2 (0.18)
2

= 4.44 × 104 NC-1

15. The quantization of electric charge is the property due to which all free charges are an
integral multiple of the basic unit of charge of an electron (or proton) represented by e.
The basic cause of quantization is that only integral no. of electrons can be transferred
from one body to another on rubbing.
16. Let two equal charges Q each, be held at A and B, where AB = 2x. C is the centre of AB,
where charge q is held, as shown in the figure below.

Net force on q is zero. So q is already in equilibrium.

For the three charges to be in equilibrium, net force on each charge must be zero.
Now, total force on Q at B is
1 Qq 1 QQ

2
+ 2
= 0
4πε0 x 4πε0
(2x)

or
Qq Q
1 1
2
= − 2
4πε0 x 4πε0 4x

or q = −
Q

Hence Proved.
→
17. Here r ⃗ 1
^ ^ ^
= ( i + 3 j + 2k)m, r2

^ ^ ^
= (3 i + 5 j + k)m

∴ ⃗
r 12 ⃗ = (3^
⃗ − r1
= r2 i + 5^ ^)
j + k

^
− (^
i + 3^
j + 2k)

^ ^ ^
= (2 i + 2 j − k)m

−−−−−−−−−−−−−
⃗
|r 12 | = √2
2
+ 2
2
+ (−1)
2
= 3m
^ ^ ^
⃗
r 22 2 i +2 j −k
r
^12 = =
⃗
r 12 ∣ 3

q1 q2
⃗ 1
F 21 = r
^12
4πε0 2
r
12

9 −6 −6
^ ^ ^
9×10 ×5×10 ×3×10 (2 i +2 j −k)
= 2
⋅
3 3

= 5 × 10-3 (2^i + 2^j − k

^ N
)

Also, F ⃗ 12
⃗
= −F 21 = -5 × 10-3 (2^i + 2^j − k
^
) N

18. a. Charge placed at the centre of the hollow sphere is -q. Hence, a charge of magnitude
+q will be induced to the inner surface. Therefore, total charge on the inner surface of
 the shell is +q. Surface charge density at the inner surface,
Total charge +q
σ = =
inner Inner surface area 4πr1
2

b. Electric field at a point lying outside the sphere at a distance rfrom the centre of the
sphere: Applying Gauss' theorem
Charge enclosed
Flux = ϕ = εo

Or, E × 4πr2 =
Q−q

ε0

Q−q
∴ E = 2
4πr ε0

Total enclosed charge

19. i. Electric flux through a Gaussian surface, ϕ = ε0

Net charge enclosed inside the shell, q = 0

Electric flux through the shell
q
∴ = 0
ε0

ii. Gauss's Law: Electric flux through a Gaussian surface is 1

ε0
times the net charge
enclosed within it.
→
Mathematically, ∮ ⃗
E ⋅ ds =
1

ε0
× q

iii. We know that electric field or net charge inside the spherical conducting shell is zero.
Hence, the force on charge Q

2
is zero.
Q
2Q(Q+ ) 2

Force on charge at A, FA = 1 2 1 3Q

2
= 2
4πε0 x 4πε0 x

Section C
20. A 0.75 g paisa coin now is made with Al only.
So, the mass of a paisa coin = 0.75g
Atomic mass of Al ≅27 a.m.u
So, number of moles in 0.75 g = 0.75

27
mole
Number of Al atoms in coin = N
= 0.75

27
× 6.022 × 10
23
atoms
Atomic number of Al = 13
∴ Number of electrons (negative charge) and protons (positive charge) = 13.

So, number of either proton or electron in a coin = 13×0.75

27
× 6.022 × 10
23

Magnitude of charge on a proton or electron = 1.6 × 10-19 C

So, total charge (either positive or negative)
23 −19

= 13×75×6.022×10

27×100
×1.6×10 C

= 3.48 × 104 C
Either positive or negative charge on a coin = 3.48 × 104 C
It concludes that even a 0.75 g Al contains an enormous amount of positive and negative
charges and equal in magnitude.
21. Let +q1 and -q2 be the initial charges on the two spheres.
a. When the two spheres attract each other,
9
F=k i.e., 0.108 = 9 × 10 ⋅
q q q q
1 2 1 2

2 2
r (0.5)

= 3 × 10-12
2

q1q2 =
0.108×(0.5)
∴ 9
9×10
 b. When the two spheres are connected by the wire, they share the charges equally.
Charge on each sphere =
q1 +(−q2 ) q −q
1 2
∴ =
2 2

Force of repulsion between them is

q −q q −q
1 2 1 2
k( )( )

F=
2 2

r2

2
9 q −q
i.e., 0.036 = 9×10
2
⋅ (
1

2
2
)
(0.5)

(q1 - q2)2 = = 4 × 10-12

2
0.036×(0.5) ×4
∴ 9
9×10

or q1 - q2 = 2 × 10-6 ...(i)
Now (q1 + q2)2 = (q1 - q2)2 + 4q1q2
= (2 × 10-6)2 + 4 × 3 × 10-12
= 16 × 10-12
∴ q1 + q2 = 4 × 10
-6 ...(ii)
On solving equations (i) and (ii), we get
-6 -6
q1 = 3 × 10 C and q2 = 10 C
which are the initial charges on the two spheres.
22. Distance between the spheres, A and B, r = 0.5 m
Initially, the charge on each sphere, q = 6.5 × 10-7 C
When an uncharged sphere is brought near the charged sphere, the charge is induced on
the uncharged sphere. Thus in the given question,
When sphere A is touched with an uncharged sphere C, the amount of charge from A
q

will transfer to sphere C. Hence, charge on each of the spheres, A and C, is .

q

When sphere C with charge q

2
is brought in contact with sphere B with charged, total
charges on the system will divide into two equal halves given as,
q
+q
2
3q
=
2 4

Each sphere will share each half. Hence, charge on each of the spheres, C and B, is .
3q

Force of repulsion between sphere A having charge and sphere B having charge
q

q 3q
× 2
3q 2 4
3q
= 2
= 2
4 4πϵ0 r 8×4πϵ0 r

2
−7
3×(6.5×10 )
= 9 × 10 9
× 2
8×(0.5)

= 5.703 × 10-3 N
Therefore, the force of attraction between the two spheres is 5.703 × 10-3 N.
23. Here, qA = 3μ C = 3 × 10-6 C ;
qB = -3μ C = - 3 × 10-6 C and r = 20 cm = 0⋅2 m
Let O be the mid-point of the line AB as shown in Fig.

Then, OA = OB = r

2
=
0⋅2

2
= 0⋅1 m
a. The electric field at point O due to qA
9 −6

EA=
q
⋅ = 9 × 10 ×
4πε0
1 A

2
3×10
2
(OA) (0⋅1)

6 1
 = 2⋅7 × 106 NC-1 (along OB)
The electric field at point O due to qB,
9 −6

EB =
q
1
⋅
4πε0
= 9 × 10 ×
A

2
3×10
2
(OB) (0−1)

6 -1
= 2⋅7 × 10 NC (along OB)
Therefore, net electric field at point O due to the charges qA and qB,
E = EA + EB = 2⋅7 × 106 + 2⋅7 × 106
= 5⋅4 × 106 N C-1 (along OB)
b. Force on a negative charge of magnitude 1⋅5 × 10-9 C placed at point O,
F = qE = 1⋅5 × 10-9 × 5⋅4 × 106
= 8⋅1 × 10-3 N (along OA)
The force on the negative charge acts in a direction opposite to that of the electric
field.
24. i. The electric field intensity due to a sheet is given by E = σ/2 ∈0.
At point A, E= 2 x 1017/2 x 8.85 x 10-12 = 1.12 x 1028 N/C
ii. At point Y, because at 50 cm, the charge sheet acts as a finite sheet and thus the
magnitude remains the same towards the middle region of the plan sheet.
25. The situation is represented in the following figure.

A and B are two parallel plates close to each other. The outer region of plate A is labelled
as I, outer region of plate B is labelled as III, and the region between the plates, A and
B, is labelled as II. Let EA and EB represents the electric field produce by plate A and B
Charge density of plate A, σ = 17.0 × 10-22 C/m2
Charge density of plate B, σ = -17.0 × 10-22 C/m2
In the regions, I and III, electric field E is zero. For region 1 E1 = EB - EA
E=σ/2 ∈o due to opposite charge densities EA is equal to zero, Similarly EB is equal to
zero
Electric field E in region II is given by the relation,
E= σ

ϵ0

ϵo = Permittivity of free space = 8.854 × 10-12 N-1C2m-2

−22

E= 17.0×10

8.854×10
−12

= 1.92 × 10-10 N/C

Therefore, the electric field between the plates is 1.92 × 10-10 N/C.
26.

Field due to a uniformly charged spherical shell is zero at all points inside the shell. So in
the given question, we have:
i. For 0 < x < a
The 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) = ⋅ 2
4πε0 x

iii. b ≤ x < ∞
Point is outside of both the spherical shells. Total effective charge at the centre equals
(Q + q)
1 q+Q
∴ E(x) = ⋅ ( )
4πε0 x2

27. a. q 1 = q2 = 6.5 × 10
−7
C

r = 50 cm = 0.50 m
1 9 2 −2
k = = 9 × 10 N m C
4πε0

F=?
According to Coulomb's law, electrostatic force of attraction is given by;
q1 q2
F = k 2
r

9 −7 −7
9×10 ×6.5×10 ×6.5×10
=
2
(0.50)

−2
F = 1.5 × 10 N

b. Now, if each sphere is charged double, and the distance between them is halved then
the force of repulsion is:
2q1 2q2
F = k ⋅ 2
r
( )
2

q1 q2
F = 16k ⋅ 2
r

−2 −2
= 16 × 1.5 × 10 = 24 × 10

F = 0.24 N
28. Suppose the three charges are placed as shown in Fig. Let the charge q be positive.

For the equilibrium of charge + q, we must have Force of repulsion F1 between + 4e and
+ q = Force of repulsion F2 between + e and + q
 or 1 4e×q 1 e×q

2
= 2
4πε0 x 4πε0
(a−x)

or 4 (a - x)2 = x2
or 2 (a - x) = ±x
∴ x= 2a

3
or 2a
As the charge q is placed between + 4e and + e, so only x = 2a

3
is possible. Hence for
equilibrium, the charge q must be placed at a distance 2a

3
from the charge + 4e.
We have considered the charge q to be positive. If we displace it slightly towards charge
e, from the equilibrium position, then F1 will decrease and F2 will increase and a net
force (F2 - F1) will act on q towards left i.e., towards the equilibrium position. Hence the
equilibrium of positive q is stable.
Now if we take charge q to be negative, the forces F1 and F2 will be attractive, as shown
in Fig.

The charge - q will still be in equilibrium at x = 2a

3
. However, if we displace charge - q
slightly towards the right, then F1 will decrease and F2 will increase. A net force (F2 - F1)
will act on - q towards the right i.e., away from the equilibrium position. So the
equilibrium of the negative q will be unstable.
29. i.

ii. a. For region II, the electric field due to sheet of charge A will be from left to right and
due to the sheet of charge B will be from right to left thus,
1
EI I = (σ1 − σ2 )
2ε0

towards right from sheet A to sheet B.

b. For region III, the electric field due to both charged sheet will be from left to right
thus,
1
EI I I = (σ1 + σ2 )
2ε0

towards the right side away from the two sheets.

30. Considering a Gaussian surface in the form of a sphere at radius r, the electric field has
the same magnitude at every point of the sphere and is directed outward. Let us consider
charge +q be uniformly distributed over a spherical shell of radius R.
'E' at any point is radially outward (if charge q is positive) and has same magnitude at all
points which lie at the same distance r from centre of spherical shell such that r > R.
Therefore, Gaussian surface is concentric sphere of radius r such that r > R.
 By Gauss' theorem
q q
∘
∮ E ⋅ dS = ⇒ ∮ EdS cos 0 =
ε0 ε0

[∵ E and dS are along the same direction]

[∵ Magnitude of E is same at every point on Gaussian surface]
q
E ⋅ ∮ dS =
ε0

2 q
E × 4πr =
ε0

1 q
⇒ E = ⋅
4πε0 r2

Now, graph is shown below.

Variation of E with r for a spherical shell in which charge q is distributed over its surface
which shows that electric field inside a spherical shell is zero.
31. a. The charge density of a long cylinder having length l and radius r1 is λ1 while other
cylinders having a similar length which surrounds the first cylinder having radius r2.
If E is the electric field in a space between the two cylinders, then as per Gauss law,
electric flux through Gaussian surface will be:
ϕ = 2πrlE

where,
r = distance of a point from the common axis of cylinders. Further, if q is the total
charge on the cylinder, so electric flux will be
q
ϕ =
ε0

where,
q = charge on inner cylinder
ε = permittivity of free space
0

Further,
λ1 l
2πrlE =
ε0

Hence, the electric field between two cylinders is

 λ1
E =
2πε0 r

b. Also, when q is the total charge on cylinder, then electric flux will be
ϕ = 2πRlE
q
ϕ =
ε0

where, q = charge, ε = permittivity of free space 0

Now, the electric field at a point outside the larger cylinder will be
(λ1 −λ2 )l
2πr2 lE =
ε0

Electric field, E
λ1 −λ2
=
2πε0 r2

So, the electric field is proportional to the difference of their charge density and
inversely proportional to the distance (radius) of the cylinder.
Section D

32.

−
→
φ1 =∮ S1
E ⃗1 ⋅ dS 1 =∮ S1
(Ex ^ ^
i ) ⋅ (dS1 i ) = ExS1
Electric flux through flat surface S2
−→
φ1 =∫ S2
⃗
E 2 ⋅ dS 2 =∫ S2
^ ^
(−Ex i ) ⋅ (−dS2 i )

=∫ S2
Ex dS2 = ExS2
Electric flux through curved surface S3

cos 90o = 0
−
→
φ3 =∫ S3
⃗
(E 3 ⋅ dS 3 ) =∫ S3
E3 dS3

∴ Net electric flux, φ = φ 1 + φ2 = Ex(S1 + S2)

But S1 = S2 = π(r × 10 ) m = πr2 ×10-4 m2
2 2
−2

∴ φ = E .2 (πr
2 × 10-4) units
x
By Gauss's law, φ = q 1

ε0

q=ε 0φ =ε 0 Ex (2πr
2
× 10
−4
)

2 −4

= 2πε 0 Ex
r
2
× 10
−4
= 4πε 0 (
Ex r ×10

2
)

2 −4

= 1 Ex r ×10

9
[ ]
9×10 2

= 5.56 Exr2 ×10-11 coulomb.

33. i.

Flux through the Gaussian surface

∅ = E.2πrl

According to Gauss’s law

q
E.2πrl =
ϵ0

∵ q = λl electric field =
λ
E =
2π∈0 r

ii. E =
λ

2πϵ0 r

2
mv
= eE
r

∴ Kinetic energy K = 1

2
mv2
= 1

2
eEr
1 λ⋅r eλ
= e =
2 2πϵ0 r 4πϵ0

iii. Kinetic energy K = eλ

4πϵ0

∴ K ∝ λ kinetic energy as a function of linear charged debnsity