1.

ELECTROSTATICS 12 PHYSICS EM RAY 1

7) The electric field lines never intersect. Justify. 16) What is meant by electrostatic energy density?
1. ELECTROSTATICS If a charge is placed in the intersection point, then it has to move in The energy stored per unit volume of space is defined as energy
two different directions at the same time, which is physically density (uE).
BOOK BACK SHORT ANSWER QUESTIONS impossible. Hence, electric field lines do not intersect.
1) What is meant by quantization of charges? 𝑼 𝟏
uE = = 𝒖𝑬 = 𝟐 𝜺𝟎 𝑬𝟐
The charge ‘q’ on any object is equal to an integral multiple of the 8) Define ‘Electric dipole’. Give the expression for the magnitude of 𝒗𝒐𝒍𝒖𝒎𝒆
fundamental unit of charge ‘e’. its electric dipole moment and the direction.
q = ne Two equal and opposite charges separated by a small distance 17) Write a short note on ‘electrostatic shielding’.
2) Write down Coulomb’s law in vector form and mention what each constitute an electric dipole. E.g. CO, water, ammonia, HCl etc., The electric field inside both hollow and solid conductor is zero (Using
term represents? Dipole moment: Gauss’s law).
The Coulomb’s law in vector form, The magnitude of the electric dipole moment is equal to the product A sensitive electrical instrument which is to be protected from
⃗ 𝟐𝟏 = 𝒌 𝒒𝟏 𝒒𝟐 of the magnitude of one of the charges and the distance between them. external electrical disturbance can kept inside the cavity of conductor.
𝒇 𝒓̂𝟏𝟐
𝒓𝟐
p = q×2a = 2qa Unit : C m This is called as electrostatic shielding.
Here, 𝑓21 is force exerted on the second charge by the first charge;
Electric dipole moment is directed from –q to +q .
𝑞1 = first charge; 𝑞2 = second charge; k = proportionality constant; 18) What is Polarisation?
9) Write the general definition of electric dipole moment for a
r = distance between the two charges; collection of point charge. Polarisation 𝑃⃗ is defined as the total dipole moment (induced) per
𝑟̂12 = is the unit vector pointing from charge 𝑞1 to 𝑞2 . The electric dipole moment for a collection of n point charges is unit volume of the dielectric. In general, the polarisation is directly
given by proportional to the strength of the external electric field.
3) What are the difference between Coulomb force and gravitational ⃗𝑷
⃗ = 𝝌𝒆 ⃗𝑬
⃗ 𝒆𝒙𝒕. Here, 𝜒𝑒 -electric susceptibility
𝑝 = ∑𝑛𝑖=1 𝑞𝑖 𝑟𝑖
force?
Here, 𝑟𝑖 is the position vector of the charge 𝑞𝑖 from the origin.
Coulomb’s force Gravitational force 19) What is dielectric field strength?
𝟏 𝓺𝟏 𝓺𝟐 𝐺𝑀1 𝑀2 The maximum electric field that the dielectric can withstand before
⃗𝑭𝟐𝟏 = 𝒓̂𝟏𝟐 𝐹= − 𝑟̂ 10) Define “Electrostatic potential”.
it breakdowns is called dielectric strength.
𝟒𝝅𝝐𝟎 𝒓𝟐 𝑟2 The electrostatic potential at a point ‘P’ is equal to the work done by
E.g. dielectric field strength of air is 3×106 V m-1.
It is act between two charges It is act between two masses an external force to bring a unit positive charge with constant velocity
from infinity to the point ‘P’ in the region of the external electric field 20) Define capacitance. Give its unit.
It is either attractive force or Only attractive force (E). The capacitance of a capacitor is defined as the ratio of the
repulsive force 𝟏 𝒒 magnitude of charge on either of the conductor plates to the potential
It is always greater in magnitude It is always smaller in magnitude
𝑽= Unit:V
𝟒𝝅𝜺𝟎 𝒓 difference existing between the conductors.
11) What is an Equi-potential Surface? 𝑄
9 2 -2 -11 2 -2
An equipotential surface is a surface on which all the points are at C = ⁄𝑉
k = 9×10 N m C G = 6.67× 10 N m kg
the same electric potential. Unit: coulomb volt-1 or farad or C 𝑉 −1 or F
Coulomb’s force depends on the Gravitational force independent of
medium the medium
12) What are the properties of an equipotential surface? 21) What is action at points or corona discharge?
When the charges are in motion, The gravitational force between two • Leakage of electric charges from the sharp end of the charged
• If two points A and B are lies on the equipotential surface, then work
yet another force (Lorentz point masses is the same whether
done to move the charge ‘q’ from A to B is zero. Since VA = VB conductor is called action at points or corona discharge.
force) comes into play in two masses are at rest or in
• The electric field must always be normal to an equipotential surface. • This is used in lightning arrester and Van Dee Graff generator.
addition to Coulomb’s force motion.
13) Give the relation between electric field and electric potential.
4) Write short notes on superposition principle. Electric field is the negative gradient of the electric potential. BOOK INTERIOR QUESTIONS
According to the superposition principle, the total force acting on a 𝒅𝑽 22) State Coulomb’s law in Electric Field.
E=- The force of attraction or repulsion between two point charges is
given charge is equal to the vector sum of forces exerted on it by all the 𝒅𝒙
other charges. directly proportional to the product of the magnitude of the two point
14) Define “electrostatic potential energy”. charges and is inversely proportional to the square of the distance
⃗ 𝒕𝒐𝒕𝒂𝒍
𝑭 ⃗ 𝟏𝟐 + 𝑭
= 𝑭 ⃗ 𝟏𝟑 + 𝑭
⃗ 𝟏𝟒 + … … 𝑭⃗ 𝟏𝒏
𝟏 If the charge ‘q2’ is brought from infinity to that point at a distance between them.
‘r’ from ‘q1’, then the work done is the product of q2 and the electric 𝟏 𝓺𝟏 𝓺𝟐
5) Define “Electric field”. potential (V)at that point. Therefore W = q2V. Actually this work done ⃗𝑭𝟐𝟏 = 𝒓̂𝟏𝟐
The electric field at the point P at a distance r from the point charge ‘q’ 𝟒𝝅𝝐𝟎 𝒓𝟐
is stored as the electrostatic potential energy ‘U’ of a system of charges
23) State the Gauss law.
is defined as the force that would be experienced by a unit positive q1 and q2 separated by a distance ‘r’.
charge. The electric flux (ΦE ) of the electric field (E) over any closed
𝟏 𝒒𝟏 𝒒𝟐 1
𝐅 1 q
U= surface is equal to (𝜀 ) times the net charge (Qencl) enclosed by the
𝟒𝝅𝜺𝟎 𝒓
⃗ =
𝐄 = r̂ Unit: N 𝐶 −1 0
𝐪𝟎 4πε0 r2 15) Define ‘Electric Flux’. surface.
The number of electric field lines crossing a given area kept normal 𝑸𝒆𝒏𝒄𝒍
6) What is mean by ‘Electric field lines’? to the electric field lines is called electric flux. 𝚽𝑬 =
𝜺𝟎
Electric field lines are the visual representation of the electric field in Φ𝐸 = 𝐸⃗ ∙ 𝐴 Φ𝐸 = 𝐸𝐴 cos 𝜃 Unit: N m2 𝐶 −1
some region of space. They form a set of continuous lines.

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 1. ELECTROSTATICS 12 PHYSICS EM RAY 2

BOOK BACK LONG ANSWER QUESTIONS Coulomb’s force Gravitational force 4) Calculate the electric field due to a dipole on its axial line and
1) Discuss the basic properties of electric charges. It is act between two charges It is act between two masses Calculate the electric field due to a dipole on its equatorial plane.
֍ Electric Charge It is either attractive force or Only attractive force
Electric charge is intrinsic and fundamental property of particles like repulsive force ELECTRIC FIELD DUE TO A DIPOLE ON ITS AXIAL LINE
‘masses’. The magnitude of fundamental unit electric charge is It is always greater in magnitude It is always smaller in magnitude for
𝑒 = 1.6 × 10−19 C small objects • Consider an electric dipole placed on the x – axis.
֍ Conservation of charges k = 9×10 9
G = 6.67× 10-11 • A point C is located at a distance of ‘r’ from the midpoint ‘O’ of the
The total charge in the universe is constant and charge can neither be N m2 C-2 N m2 kg-2 dipole along the axial line.
created nor be destroyed. In any physical process, the net change in Coulomb’s force depends on the Gravitational force independent of the
charge will always be zero. medium medium
֍ Quantization of charges When the charges are in motion, The gravitational force between two
The total charge ‘q’ of any object is equal to an integral multiple of yet another force (Lorentz force) point masses is the same whether two
this fundamental unit of electric charge ‘e’. comes into play in addition to masses are at rest or in motion.
𝑞 = 𝑛𝑒 Coulomb’s force
Here, n is an integer.( 0, ±1 , ±2, ±3, ±4 … … . )
𝑒 = 1.6 × 10−19 𝐶 3) Define ‘Electric Field’ and discuss its various aspects.
The electric field at the point P at a distance’ r’ from the point charge
2) Explain in detail Coulomb’s law and its various aspects. ‘q’ is the force experienced by a unit charge(q0) and is given by
֍ Coulomb’s law The electric field at a point C:
⃗𝑭 𝟏 𝒒𝒒𝟎 𝟏 𝒒
The force of attraction or repulsion between two charges (q1, q2) is ⃗ =
𝑬 = 𝟒𝝅𝜺 𝒓̂ = 𝟒𝝅𝜺 𝒓̂
𝒒𝟎 𝒓𝟐 𝒓𝟐 1 𝑞
directly proportional to product of charges and is inversely proportional
𝟎 𝟎
• Due to +q is, 𝐸⃗+ = 𝑝̂ (𝑎𝑙𝑜𝑛𝑔 𝐵𝐶) →→→ ( 1 )
Electric field is a vector quantity. Unit = N C-1 or V m-1 4𝜋𝜀0 (𝑟−𝑎)2
to square of the distance (r) between them.
1 𝑞
⃗ 𝟐𝟏 =
𝟏 𝓺𝟏 𝓺𝟐 ֍ Important aspects of electric field • Due to –q is, 𝐸⃗− = − 4𝜋𝜀 (𝑟+𝑎)2
𝑝̂ ( 𝑎𝑙𝑜𝑛𝑔 𝐶𝐴) →→→ ( 2 )
𝑭 𝒓̂𝟏𝟐 0
𝟒𝝅𝝐𝟎 𝒓𝟐 • q is positive -The electric field points radially away from the source
֍ Important aspects of Coulomb’s law charge. 𝐸⃗ tot = 𝐸⃗+ + 𝐸⃗− →→→ ( 3 )
⃗𝑭𝟐𝟏 = Force acting on the second charge by the first charge. 𝒓̂𝟏𝟐 = unit q is negative - The electric field points radially towards the source Substitute (1) and (2) in (3).
vector directed from the first charge (q1) to second charge (q2). charge.
1 𝑞 1 𝑞
⃗ 𝟏𝟐 = Force acting on the first charge (q1) by the second charge (q2), its
𝑭 • 𝐹 = 𝐸⃗ 𝑞0 . This is Coulomb’s law in terms of electric field. ▪ 𝐸⃗ tot = 𝑝̂ − 𝑝̂
4𝜋𝜀0 (𝑟−𝑎)2 4𝜋𝜀0 (𝑟+𝑎)2
direction is from second charge to first charge(- 𝒓̂𝟏𝟐 ). • From the above equation the electric field is independent of the test charge
1 1 1
and it is depends on source charge. ▪ 𝐸⃗𝑡𝑜𝑡 = [(𝑟−𝑎)2 − (𝑟+𝑎)2
] 𝑝̂
4𝜋𝜀0
The value of
𝟏
= 9 × 109 N 𝑚2 𝐶 −2 . • When the distance from the source charge increases then the electric field
𝟒𝝅𝝐𝟎 𝑞 4𝑟𝑎
𝜀0 = 8.854 × 10 −12 2
𝐶 𝑁 −1 −2
𝑚 .
is decreases in magnitude. ▪ 𝐸⃗𝑡𝑜𝑡 = [(𝑟 2 ] 𝑝̂ →→→ (4)
• The test charge is made sufficiently small such that it will not modify the 4𝜋𝜀0 − 𝑎2 )2

electric field of the source charge. ( r >> a), then the above equation can be written as,
▪ The Coulomb’s law can be written as 𝐹12 = − 𝐹21 • The expression 𝐹 = 𝐸⃗ 𝑞0 is valid only for point charges. For continuous
▪ From the above equation we came to conclusion the electrostatic force 1 4𝑎𝑞
obeys Newton’s Third law.
and finite size charge distributions, integration techniques must be used. • 𝐸⃗𝑡𝑜𝑡 = ( 𝑟 3 ) 𝑝̂
4𝜋𝜀0
▪ 𝑰𝒇 𝓺𝟏 = 𝓺𝟐 = 1C & r = 1 m. Uniform electric field Non-uniform electric field
9×109 ×1×1 • Uniform electric field will have the • Non-uniform electric field will • 𝑝 = 2𝑞𝑎𝑝̂
Then |𝐹| = = 9 × 109 𝑁 same direction and constant have different directions or
12 𝟏 ⃗
𝟐𝒑
▪ The force between two point charges in a medium other than vacuum is magnitude at all points in space. different magnitudes or both at • ⃗𝑬
⃗ 𝒕𝒐𝒕 =
𝟒𝝅𝜺𝟎 𝒓𝟑
always less than that in vacuum. different points in space.
֍ Note
• 𝜖 = 𝜖0 𝜖𝑟 , • If two oppositely charged plates are • The electric field created by the
• 𝜖0 = permittivity of free space or vacuum; placed very close to each other, may point charge is basically a non- The direction of ⃗𝑬𝒕𝒐𝒕 is in the direction of 𝑝
• 𝜖𝑟 = Relative permittivity of the medium; produce uniform electric field. uniform electric field.
• 𝜖 = permittivity of the medium.
(For vacuum or air 𝜖𝑟 = 1, for all other medium (𝜖𝑟 > 1 ))
▪ The expression for Coulomb’s force is true only for point charges.

• Coulomb’s law has same structure as Newton’s law of gravitation. But

there is some important difference between these two laws.

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 1. ELECTROSTATICS 12 PHYSICS EM RAY 3
5) Derive and expression for the torque experienced by a dipole due to ⃗⃗⃗⃗ = 𝑑𝑟𝑟̂ ) and ((𝑟 ̂ ∙ 𝑟 ̂ ) = 1)
(𝑑𝑟
ELECTRIC FIELD DUE TO A DIPOLE ON EQUITORIAL a uniform electric field 1 𝑟 1
PLANE 𝑉= − 𝑞 ∫∞ 2 𝑟̂ ∙ 𝑑𝑟𝑟̂
4𝜋𝜀0 𝑟
𝟏 𝒒
𝑽=
𝟒𝝅𝜺𝟎 𝒓

7) Derive an expression for electrostatic potential due to an electric

dipole.

• Consider an electric dipole of dipole moment 𝑝 placed in a uniform

electric field 𝐸⃗ .
• The force acting on +q charge is 𝐹+ = 𝑞𝐸⃗
• The point P is located at a distance ‘r’ from the midpoint of the dipole.
• Consider a point C at a distance ‘r’ from the midpoint ‘O’ of the • And the force acting on the –q charge is 𝐹− = −𝑞𝐸⃗ Let 𝜃 be the angle between the line OP and dipole axis AB.
dipole on the equatorial plane. • Resultant force acting on the dipole is zero. • Let r1 and r2 be the distance of point P from +q and –q.
1 𝑞 1 𝑞
• V1 = 4𝜋𝜀 𝑟 →→→( 1 ) V2 = − 4𝜋𝜀 𝑟 →→→ ( 2 )
• The point ‘C’ is equi-distant from +q and –q, the • Due to the couple, the dipole experience a torque, this torque tends to 0 1 0 2
𝑽 = 𝑽𝟏 + 𝑽𝟐 →→→ ( 3 )
Magnitude of the electric fields of +q and –q are same. align (rotate) the dipole in the direction of external electric field. Substitute (1) and (2) in (3),
1 𝑞 1 𝑞
▪ ⃗⃗⃗⃗⃗ × (−𝑞𝐸⃗ )) + (𝑂𝐵
𝜏 = (𝑂𝐴 ⃗⃗⃗⃗⃗ × 𝑞𝐸⃗ )→→→ ( 1 ) 𝑉 = (4𝜋𝜀 𝑟 ) + (− 4𝜋𝜀 𝑟 )
• 𝐸⃗+ and 𝐸⃗− are resolved into two components, 1
0 1
1 1
0 2

▪ ⃗⃗⃗⃗⃗ ||−𝑞𝐸⃗ | sin 𝜃 + |𝑂𝐵

𝜏 = |𝑂𝐴 ⃗⃗⃗⃗⃗ ||𝑞𝐸⃗ | sin 𝜃 →→→ (2) 𝑉 = 4𝜋𝜀 𝑞 (𝑟 − 𝑟 ) →→→ ( 4 )
0 1 2
• The perpendicular components |𝐸⃗+ | sin 𝜃 and |𝐸⃗− | sin 𝜃 are Apply the cosine law for triangle BOP, (and use binomial theorem)
▪ (Since, |𝑂𝐴 ⃗⃗⃗⃗⃗ | = 𝑎, |𝑞𝐸⃗ | = |−𝑞𝐸⃗ | , 𝑎 × 𝑏⃗ = 𝑎𝑏𝑠𝑖𝑛𝜃 )
⃗⃗⃗⃗⃗ | = |𝑂𝐵
oppositely directed and cancel each other. 𝑟12 = 𝑟 2 + 𝑎2 − 2𝑟𝑎 cos 𝜃
▪ 𝜏 = 𝑎𝑞𝐸 sin 𝜃 + 𝑎𝑞𝐸 sin 𝜃 𝟏 𝟏 𝒂
= 𝒓 ( 𝟏 + 𝒓 𝐜𝐨 𝐬 𝜽) →→→ ( 5 )
• The magnitude of the total electric field at point C is the sum of the 𝒓𝟏
▪ 𝜏 = 2𝑎𝑞𝐸 sin 𝜃 [ p = q 2a] 𝑟22 = 𝑟 2 + 𝑎2 − 2𝑟𝑎 𝑐𝑜𝑠(180 − 𝜃 )
parallel components of ⃗⃗⃗𝐸+ and 𝐸⃗− , Its direction is along - 𝑝̂ . Similarly
𝜏𝐸 = 𝑝𝐸 sin 𝜃 𝟏 𝟏 𝒂
= 𝒓 ( 𝟏 − 𝒓 𝐜𝐨 𝐬 𝜽) →→→ ( 6 )
• 𝐸⃗𝑡𝑜𝑡 = −|𝐸⃗+ | cos 𝜃 𝑝̂ − |𝐸⃗− | cos 𝜃 𝑝̂ → ( 1 ) 𝒓𝟐
▪ ⃗ × ⃗𝑬
⃗ = 𝒑
𝝉 1 1 𝑎 1 𝑎
• |𝐸⃗+ | = |𝐸⃗− | 𝑉= 𝑞 ( ( 1 + cos 𝜃) − ( 1 − cos 𝜃) )
6) Derive an expression for electrostatic potential due to a point 4𝜋𝜀0 𝑟 𝑟 𝑟 𝑟
1 𝑞 1 𝑞2𝑎
• |𝐸⃗+ | = |𝐸⃗− | = →→→ ( 2 ) charge. 𝑉= cos 𝜃
4𝜋𝜀0 (𝑟 2 + 𝑎2 ) 4𝜋𝜀0 𝑟 2
1 2𝑞 cos 𝜃 𝟏 𝒑𝐜𝐨 𝐬 𝜽
• 𝐸⃗𝑡𝑜𝑡 = - 4𝜋𝜀 (𝑟 2 + 𝑎2 ) 𝑝̂
→→→ ( 3 ) 𝑽= ( ) p = q 2a
0 𝟒𝝅𝜺𝟎 𝒓𝟐
• From the first diagram, cos 𝜃 can be written as,
1 𝑝∙𝑟̂
▪ cos 𝜃 =
𝑎
→→→ ( 4 ) In vector notation 𝑉 =
√(𝑟 2 + 𝑎2 ) 4𝜋𝜀0 𝑟 2
1 2𝑞𝑎
SPECIAL CASES
▪ 𝐸⃗𝑡𝑜𝑡 = - 𝑝̂ →→→ (5) 𝑝 = 2𝑞𝑎𝑝̂ Consider a positive charge q kept fixed at the origin. Let ‘P’ be a point θ = 0° 1 𝑝 point P lies on the axial line of the
4𝜋𝜀0 (𝑟 2 + 𝑎2 )3⁄2 𝑉=
at a distance r from the charge q. 4𝜋𝜀0 𝑟 2 dipole on the side of +q
𝟏 ⃗
𝒑 𝑟
• ( r >> a ) → ⃗𝑬𝒆𝒒𝒖𝒂 = - 𝟒𝝅𝜺 𝟑
⃗⃗⃗⃗
𝑉 = ∫ (−𝐸⃗ ) ∙ 𝑑𝑟
𝟎𝒓 ∞
From the definition electric field due to positive point charge at a 𝜃 = 180° 1 𝑝 point P lies on the axial line of the
𝑉= −
Note: distance ‘r’ is 4𝜋𝜀0 𝑟 2 dipole on the side of –q
The direction of resultant electric field is opposite to the direction of 1 𝑞
electric dipole moment. 𝐸⃗ = 𝑟̂ 𝜃 = 90° 𝑉=0 point P lies on the equatorial line of
4𝜋𝜀0 𝑟 2
1 𝑟 1
⃗⃗⃗⃗
𝑉 = − 4𝜋𝜀 𝑞 ∫∞ 𝑟 2 𝑟̂ ∙ 𝑑𝑟 the dipole
0

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 1. ELECTROSTATICS 12 PHYSICS EM RAY 4
8) Obtain an expression for potential energy due to a collection of • This torque(𝜏𝐸 ) tend to align the dipole along the electric field(𝐸⃗ ).
three point charges which are separated by finite distances.
• The workdone by the external torque to rotate the dipole from angle
𝜃 ′ 𝑡𝑜 𝜃 at constant angular velocity,
𝜃
• 𝑊 = ∫𝜃′ 𝜏𝑒𝑥𝑡 𝑑𝜃

• But, |𝜏𝑒𝑥𝑡 | = |𝜏𝐸 | = |𝑝 × 𝐸⃗ | = pE sin 𝜃

𝜃
• W = ∫𝜃′ 𝑝𝐸𝑠𝑖𝑛𝜃𝑑𝜃 = 𝑝𝐸(− cos 𝜃)𝜃𝜃′
• 𝑊 = 𝑈 = −𝑝𝐸 (cos 𝜃 − cos 𝜃 ′ )
Three charges are arranged in the following configuration as shown in the
following. • According Work – Energy principle, This work is stored as
• Bring the charge q1 from infinity to the point A requires no work potential energy in the system.
• To bring the second charge q2 to the point B, work must be done Φ𝐸 = ∫ 𝐸⃗ ∙ 𝑑𝐴 + ∫ 𝑡𝑜𝑝
𝐸⃗ ∙ 𝑑𝐴 + ∫ 𝐸⃗ ∙ 𝑑𝐴
′ ° 𝑐𝑢𝑟𝑣𝑒𝑑 𝑏𝑜𝑡𝑡𝑜𝑚
against the electric field created by the charge q1. So the workdone • If the initial angle 𝜃 = 90 and is taken as reference point, then 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
on the charge q2 is • On the curved surface 𝐸⃗ and 𝐴 are parallel to each other .
• 𝑈 = −𝑝𝐸 cos 𝜃 ⃗ ∙ ⃗𝑬
𝑼 = −𝒑 ⃗
 𝑊 = 𝑞2 𝑉1𝐵 𝐸⃗ ∙ 𝑑𝐴 = 𝐸𝑑𝐴. ( θ = 0° then cos 0° = 1 )
 𝑈𝐼 = 𝑞2 𝑉 =
1 𝑞1 𝑞2 • If 𝜃 = 0° → U = - pE ; If 𝜃 = 180° → U = +PE • For the top and bottom surfaces,𝐸⃗ and 𝐴 are perpendicular to each other.
4𝜋𝜀0 𝑟
 To bring the charge q3 to the point C, • If 𝜃 = 90° → U = 0 𝐸⃗ ∙ 𝑑𝐴 = 0 ( 𝜃 = 90° and hence cos 90° = 0 )
 𝑊 = 𝑞3 (𝑉1𝐶 + 𝑉2𝐶 ) • substituting these values in the above equation,
10) Obtain Gauss’s law from Coulomb’s law. • Φ𝐸 = ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝐸𝑑𝐴
Gauss’s law can be proven from Coulomb’s law if it assumed, in 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
1 𝑞1 𝑞3 𝑞2 𝑞3 addition that the electric field obeys the super position principle.
 𝑈𝐼𝐼 = (𝑟 + ) • From the definition of Gauss’s law, Φ𝐸 =
𝑄𝑒𝑛𝑐𝑙
4𝜋𝜀0 13 𝑟23 Consider a point charge ‘+q’. From this point charge at a distance ‘r’, 𝜀0
 𝑈 = 𝑈𝐼 + 𝑈𝐼𝐼 we consider a point c. 𝑄𝑒𝑛𝑐𝑙
▪ E=𝑞
𝐹 ▪ ∫ 𝑐𝑢𝑟𝑣𝑒𝑑 𝐸𝑑𝐴 =
𝑠𝑢𝑖𝑟𝑓𝑎𝑐𝑒 𝜀0
1 𝑞 𝑞2 𝑞1 𝑞3 𝑞2 𝑞3 0
 𝑈= ( 𝑟1 + + )
4𝜋𝜀0 𝑟13 𝑟23
12
▪ (𝑞0 = 1 C ; 𝐹 =
1 𝑞𝑞0
𝑇ℎ𝑖𝑠 𝑖𝑠 𝐶𝑜𝑢𝑙𝑜𝑚𝑏 ′ 𝑠 𝑙𝑎𝑤) • The linear charge density (Charge per unit length)
Note 4𝜋𝜀0 𝑟 2
𝑄𝑒𝑛𝑐𝑙
▪ The stored potential energy U is equal to the total external work done • 𝜆=
1 𝑞
𝐸⃗ =
𝐿
to assemble the three charges at the given locations. ▪ Therefore, 𝑟̂
▪ The electrostatic potential energy is independent of the manner in 4𝜋𝜀0 𝑟2 • The curved surface area = 2𝜋𝑟𝐿
which the configuration of charges is arrived at.
▪ Φ𝐸 = ∮ 𝐸⃗ ∙ 𝑑𝐴 • E2𝜋𝑟𝐿 =
𝜆𝐿
𝜀0
9) Derive an expression for electrostatic potential energy of the 1 𝑞
▪ Φ𝐸 = 𝑟̂ ∮ 𝑑𝐴 𝟏 𝝀
dipole in a uniform electric field. 4𝜋𝜀0 𝑟 2 • 𝑬 = 𝟐𝝅𝜺
𝟎 𝒓
1 𝑞 2 2
▪ Φ𝐸 = 4𝜋𝑟 (𝑟̂ = 1; ∮ 𝑑𝐴 = 4𝜋𝑟 )
4𝜋𝜀0 𝑟 2
1 𝜆
𝑞 The above equation in vector notation, 𝐸⃗ = 2𝜋𝜀 𝑟̂
▪ Φ𝐸 = This is Gauss’s law. 0 𝑟
𝜀0
𝐼𝑓 𝜆 > 0 𝑡ℎ𝑒𝑛 𝐸 𝑝𝑜𝑖𝑛𝑡 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑎𝑛𝑑 𝑖𝑓 𝜆 < 0 𝑡ℎ𝑒𝑛 𝐸 𝑝𝑜𝑖𝑛𝑡 𝑖𝑛𝑤𝑎𝑟𝑑
So we can able to derive Gauss’s law from Coulomb’s law.
12) Obtain the expression for electric field due to a charged infinite
11) Obtain the expression for electric field due to an infinitely long plane sheet.
charged wire.
• Consider an infinitely long straight wire having uniform linear charge
density 𝜆. Let P be a point located at a perpendicular distance ‘r’ from
the wire.
• Consider an electric dipole of dipole moment 𝑝 placed in a uniform • Let us choose a cylindrical Gaussian surface of radius r and length L as
show in the in the figure.
electric field 𝐸⃗ .
• A dipole experience a torque when kept in an uniform electric field 𝐸⃗ . • The total electric flux in this closed surface is calculated as follows.
(𝜏𝐸 = 𝑝 × 𝐸⃗ ) 𝒑 = 𝒒𝟐𝒂 Φ𝐸 = ∮ 𝐸⃗ ∙ 𝑑𝐴

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 1. ELECTROSTATICS 12 PHYSICS EM RAY 5
• Consider an infinite plane sheet of charges with uniform surface 14) Discuss the various properties of conductors in electrostatic • When the grounding wire is removed from the conductor, the positive
𝑄𝑒𝑛𝑐𝑙 equilibrium. charges remain near the charged rod (Observe the figure ‘c’)
charge density(𝜎 = ). That is charge per unit area. Let P be a
𝐴 ֍ Electrostatic equilibrium
point at a distance of r from the sheet . When there is no external field, the free electrons are in continuous
• A cylindrical shaped Gaussian surface of length 2r and area A of the • Now the charged rod is taken away from the conductor. As soon as the
random motion in all directions. As a result, there is no net motion of
flat surfaces is chosen charged rod is removed, the positive charge gets distributed uniformly on
electrons along any particular direction which implies that the conductor
⃗⃗⃗⃗⃗ = 0) the surface of the conductor. By this process, the neutral conducting
• E- perpendicular to the curved surface (∫𝑐𝑢𝑟𝑣𝑒𝑑 𝐸⃗ ∙ 𝑑𝐴 is in electrostatic equilibrium. The time taken by a conductor to reach
sphere becomes positively charged.
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 electrostatic equilibrium is in the order of 10-16 second.( that is almost
• E- parallel to the surface areas at P and 𝑃′ instantaneous)
• Φ𝐸 = ∮ 𝐸⃗ ∙ ⃗⃗⃗⃗⃗
𝑑𝐴 16) Explain dielectrics in detail and how an electric field is induced
֍ Properties
inside a dielectric.
▪ Φ𝐸 = ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝐸⃗ ∙ 𝑑𝐴⃗⃗⃗⃗⃗ + ∫ 𝐸⃗ ∙ 𝑑𝐴
⃗⃗⃗⃗⃗ + ∫ ′ 𝐸⃗ ∙ 𝑑𝐴
⃗⃗⃗⃗⃗ • The electric field is zero everywhere inside the conductor. This is true
𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝑃 𝑃 ֍ Dielectric
regardless of whether the conductor is solid or hollow.
• Apply the Gauss’s law for this cylindrical surface, A dielectric is non-conducting material and has no free electrons. The
• There is no net charge inside the conductors. The charges must reside
𝑄 electrons in a dielectric are bound within the atoms. Examples Ebonite,
• Φ𝐸 = ∫𝑃 𝐸𝑑𝐴 + ∫𝑃′ 𝐸𝑑𝐴 = 𝑒𝑛𝑐𝑙 only on the surface of the conductors.
𝜀 0 glass, mica, wood, rubber, plastic etc., A dielectric is made up of either
• E-is uniform at p and 𝑃′ • The electric field outside the conductor is perpendicular to the surface
𝜎 polar molecules or non-polar molecules.
• 𝑄𝑒𝑛𝑐𝑙 = 𝜎𝐴, of the conductor and has a magnitude of 𝜀 where 𝜎 is the surface charge
0
𝜎𝐴
• 2𝐸 ∫𝑃 𝑑𝐴 = 𝜀 density at that point. ֍ Inducing of electric field inside the dielectric
𝜎𝐴
0
• The electrostatic potential has the same value on the surface and inside In the case of conductor, when an external electric field is applied on
• 2𝐸𝐴 = (Hence, ∫𝑃 𝑑𝐴 = 𝐴) of the conductor. a conductor, the charges are aligned in such a way that an internal electric
𝜀0
𝝈 field is created which cancels the external electric field. But, in the case
• 𝑬 = 𝟐𝜺
𝟎 15) Explain the process of electrostatic induction. of a dielectric, which has no free electrons, the external electric field only
𝜎
• In vector notation, 𝐸⃗ = 2𝜀 𝑛̂ ֍ Electrostatic induction realigns the charges so that an internal electric field is produced.
0
If 𝜎 > 0 𝑡ℎ𝑒𝑛 𝐸 𝑝𝑜𝑖𝑛𝑡 𝑜𝑢𝑡𝑤𝑎𝑟𝑑 𝑎𝑛𝑑 𝑖𝑓 𝜎 < 0 𝑡ℎ𝑒𝑛 𝐸 𝑝𝑜𝑖𝑛𝑡 𝑖𝑛𝑤𝑎𝑟𝑑 The magnitude of the internal electric field is smaller than that of
external electric field. Therefore the net electric field inside the dielectric
13) Obtain the expression for electric field due to a uniformly is not zero but is parallel to an external electric field with magnitude less
charged spherical shell. than that of the external electric field.
Consider a uniformly charged spherical shell or radius R and total
charge Q as shown in the figure.

At a point outside At a point on the At a point inside the

the shell surface spherical shell Charging without actual contact is called electrostatic induction
(r>R) (r=R) (r<R) • Consider an uncharged conducting sphere at rest on an insulating stand.
Suppose a negatively charged rod is brought near the conductor without
Applying Gauss law, Applying Gauss law, Applying Gauss law, touching it. In the figure ‘a’. The negative charge of the rod repels the
𝑄 𝑄 𝑄
∮ 𝐸⃗ ∙ 𝑑𝐴 = ∮ 𝐸⃗ ∙ 𝑑𝐴 = ∮ 𝐸⃗ ∙ 𝑑𝐴 = electrons in the conductor to the opposite side. As a result, positive
𝜀0 𝜀0 𝜀0
charges are induced near the region of the charged rod while negative
Gaussian surface Gaussian surface Gaussian surface charges on the farther side.
encloses charge Q encloses charge Q encloses no charge
• Now the conducting sphere is connected to the ground through a
∮𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑑𝐴 =4𝜋𝑟 2 ∮𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑑𝐴 ∮𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑑𝐴 =4𝜋𝑟 2 conducting wire. This is called grounding. Grounding removes the
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
=4𝜋𝑅 2 electron from the conducting sphere. Note that positive charges will not
1 𝑄 1 𝑄 𝐸=0 ( since Q = 0) flow to the ground because they are attracted by the negative charges of
𝐸⃗ = 𝑟̂ 𝐸⃗ = 𝑟̂
4𝜋𝜀0 𝑟 2 4𝜋𝜀0 𝑅 2 the rod.(Observe the figure ‘b’)

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 1. ELECTROSTATICS 12 PHYSICS EM RAY 6
֍ Experimental setup 19) Explain detail the effect of a dielectric placed in a parallel plate 20) Derive an expression for resultant capacitance, when capacitors
Let us consider a rectangular dielectric slab placed between two capacitor are connected in series and parallel
oppositely charged plates a shown in the following figure. • Consider a capacitor with two parallel plates each of cross-sectional area
The uniform electric field between the plates acts as an external A and are separated by a distance ‘d’. The capacitor is charged by a capacitors in series Capacitors in parallel
electric field 𝐸⃗𝑒𝑥𝑡 which polarizes the dielectric placed between plates. battery of voltage V0 and the charge stored is Q0.
The positive charges are induced on one side surface and negative charges • Electric field of the capacitor without dielectric is- 𝐸0 • Consider three capacitors of • Consider three capacitors of
are induced on the other side of surface. • The capacitance of the capacitor without dielectric is-𝐶0 = 𝑉0
𝑄 capacitance 𝐶1 , 𝐶2 𝑎𝑛𝑑 𝐶3 capacitance 𝐶1 , 𝐶2 𝑎𝑛𝑑 𝐶3
0
Inside the dielectric, the net charge is zero even in small volume. From
the diagram the dielectric in the external field is equivalent to two • The energy stored in the capacitor before the insertion of a dielectric is Connected in series with a battery Connected in parallel with a
1 𝑄02 battery of voltage V
Oppositely charged sheets with the surface charge densities +σb and -σb. given by 𝑈0 = of voltage V
2 𝐶0
These charges are called bound charges. They are not free to move like
• 𝜀𝑟 -is the relative Permeability of the dielectric or dielectric constant
free electrons in conductors.
For example, the charged balloon after rubbing sticks onto a wall. The
When the capacitor is When the capacitor is
reason is that the negatively charged balloon is brought near the wall, it disconnected from the battery connected to the battery
polarizes opposite charges on the surface of the wall, which attracts the
balloon.
17) Obtain the expression for capacitance for a parallel plate
capacitor.
▪ Capacitor is device used to store electric charge and electrical
energy.
• A simple capacitor consists of two parallel metal plates of area (A)
separated by a small distance (d)
𝑄
• Surface charge density (𝜎 = ),
𝐴
𝜎
• Electric field between two plates is (𝐸 = 𝜀 ) 𝑉 = 𝑉1 + 𝑉2 + 𝑉3 𝑄 = 𝑄1 + 𝑄2 + 𝑄3
𝑄
0
Since, 𝜀𝑟 > 1, E < E0; V < V0 V0 – constant; E0 - remain constant
• 𝐸= 𝑄 𝑄
𝐴𝜀0 𝐸0 𝑉0 𝐶= 𝐶=
𝑄𝑑 𝐸= 𝑉= 𝑉 𝑉
• V =Ed = 𝜀𝑟 𝜀𝑟
𝐴𝜀0
• 𝐶=𝑉
𝑄 𝑄 𝑄 𝑄 𝑄 = 𝐶1 𝑉 + 𝐶2 𝑉 + 𝐶3 𝑉
Q0 - remain constant Q0 - increased 𝑉1 = ; 𝑉2 = ; 𝑉3 =
𝑄 𝐴𝜀0 𝐶1 𝐶2 𝐶3
• 𝐶 = 𝑄𝑑 = 𝑄 = 𝜀𝑟 𝑄0 Q > Q0
⁄𝐴𝜀 𝑑
0 1 1 1 𝐶𝑃 𝑉 = 𝐶1 𝑉 + 𝐶2 𝑉 + 𝐶3 𝑉
𝛆𝟎 𝐀 𝑉 = 𝑄( + + )
• 𝑪= Since, V < V0, we have C > C0 Since, 𝑄 > 𝑄0 , we have C > C0 𝐶1 𝐶2 𝐶3
𝒅

𝑄0 𝑄0 𝑄 𝑄0
𝐶= = 𝜀𝑟 = 𝜀𝑟 𝐶0 𝐶= = 𝜀𝑟 = 𝜀𝑟 𝐶0 We can write, 𝑄 = 𝐶𝑃 𝑉
𝑉 𝑉0 𝑉0 𝑉0 𝑄
But,𝑉 = 𝐶
𝑆
𝜖𝐴 𝜖𝐴
18) Obtain the expression for energy stored in the parallel plate 𝐶= 𝐶=
𝑑 𝑑
capacitor.
• Capacitor is device used to store electric charge and electrical energy. Before the insertion of a 1 𝟏 𝟏 𝟏 𝟏 𝑪𝑷 = 𝑪𝟏 + 𝑪𝟐 + 𝑪𝟑
• To transfer the charge work is done by the battery. 𝑈0 = 𝐶 𝑉2 = + +
1 𝑄02 2 0 0 𝑪𝑺 𝑪𝟏 𝑪𝟐 𝑪𝟑
• This workdone is stored as electrostatic potential energy in the dielectric: 𝑈0 =
2 𝐶0
capacitor.
𝑄
• 𝑑𝑊 = 𝑉 𝑑𝑄 Where 𝐶 = 𝑉 After the insertion of a 𝑈 = 𝑈0
𝑄 𝑼𝟎
• 𝑑𝑊 = 𝐶 𝑑𝑄 dielectric: 𝑼= 𝜺𝒓 U > U0.
• The total workdone to charge a capacitor is,
• 𝑊 = ∫0𝑄 𝑑𝑊 U < U0. The energy density 𝒖 = 𝟐 𝜺𝑬𝟐𝟎
𝟏

𝑄𝑄 𝑄2
• 𝑊 = ∫0 𝐶
𝑑𝑄 = 2𝐶
𝑄2
• 𝑈𝐸 = (since, 𝑊 = 𝑈𝐸 )
2𝐶
1
• 𝑈𝐸 = 2 𝐶𝑉 2 since, 𝑄 = 𝐶𝑉

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 1. ELECTROSTATICS 12 PHYSICS EM RAY 7
21) Explain in detail how charges are distributed in a conductor and ֍ Working ֍ Working
the principle behind the lightning conductor. When a negatively charged cloud ▪ Due to the high electric field near comb D, air between the belt and
is passing above the building, it comb D gets ionized by the action of points.
induces a positive charge on the spike. ▪ positive charges are pushed towards the belt and negative charges are
Since the induced charge density on attracted towards comb D.
thin sharp spike is large, it results in a ▪ The positive charges stick to the belt and move up.
corona discharge. This positive ▪ When the positive charges reach the comb E, a large amount of
charge ionizes the surrounding air negative and positive charges are induced on either side of comb E due
which in turn neutralizes the negative to electrostatic induction.
charge in the cloud. ▪ As a result, the positive charges are pushed away from the comb E and
The negative charge pushed to the they reach the outer surface of the sphere.
Consider two conducting spheres A and B of radii r1 and r2 respectively spikes passes through the copper rod ▪ Where the potential difference of the order of 107 volt is produced
connected to each other by a thin conducting wire. The distance between and is safely diverted to the earth. which is the limiting value.
the spheres is much greater than the radii of either sphere. ֍ Note ֍ Leakage of electric charges and reducing
• If a charge Q is introduced into any one of the spheres. The lightning arrester does not The leakage of charges can be reduced by enclosing the machine in a
• Let q1 - charge residing on the surface of sphere A stop the lightning; rather it diverts the gas filled steel chamber at very high pressure.
q2- charge residing on the surface of sphere B lightning to the ground safely.
𝑉𝐴 =
1 𝑞1 ֍ Application
4𝜋𝜀0 𝑟1 The high potential is used to accelerate the positive ions (protons and
1 𝑞2
𝑉𝐵 = 4𝜋𝜀 22) Explain in detail the construction and working of a Van de Graaff deuterons) for nuclear disintegrations and other applications.
0 𝑟2
generator.
֍ Since the spheres are connected by the conducting wire, the surfaces
֍ Principle
of both the spheres together form an equipotential surface.
Electrostatic induction and Action at points
𝑉𝐴 = 𝑉𝐵
֍ Input = 104 volt
1 𝑞1 1 𝑞2 ֍ Output = 107 volt
= BOOK INTERIOR QUESTIONS
4𝜋𝜀0 𝑟1 4𝜋𝜀0 𝑟2 ֍ Structure 1. Write the application of capacitor. Mention its disadvantages.
𝑞1 𝑞2 • A - hollow spherical conductor is fixed on the insulating stand ֍ Application
= • B ,C – pulley • Flash capacitor used in camera
𝑟1 𝑟2 • During cardiac arrest, defibrillator is used to retrieve the normal heart
𝑞 • A belt made up of insulating materials like silk or rubber runs over
From the definition surface charge density, 𝜎 = 𝐴 function.
both pulleys.
That is 𝑞 = 𝜎 𝐴 • E and D - comb shaped metallic conductors. • Capacitors are used in the ignition system of automobile engines to
𝑞1 = 4𝜋𝑟12 𝜎1 • D - is maintained at 104 volt by a power supply. eliminate sparking.
𝑞2 = 4𝜋𝑟22 𝜎2 • The upper comb E is connected to the inner side of the hollow • Capacitors are used to reduce power fluctuations in power supplies and
metal sphere. to increase
4𝜋𝑟12 𝜎1 4𝜋𝑟22 𝜎2 • the efficiency of power transmission.
= • ֍ Disadvantages
𝑟1 𝑟2
• Even after the battery or power supply is removed, the capacitor stores
𝝈𝟏 𝒓𝟐 = 𝝈𝟏 𝒓𝟐 charges and energy for some time. For example, if the TV is switched
𝝈 𝒓 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 off, it is always advisable to not to touch the back side of the TV panel.
֍ Note
From the above equation the surface charge density 𝜎 is inversely 2. Define polar and non-polar molecules.
proportional to the radius of the sphere. For a smaller radius, the charge Polar molecules Non-polar molecules
density will be larger and vice versa.
In polar molecules, the centers A non-polar molecule is one in
LIGHTNING ARRESTER OR LIGHTNING CONDUCTOR of the positive and negative charges which centers of positive and
This is a device used to protect tall buildings from lightning strikes. are separated even in the absence of negative charges coincide. As a
֍ Principle an external electric field. They result, it has no permanent dipole
It works on the principle of action at points or corona discharge.
֍ Structure have a permanent dipole moment. moment.
This device consists of a long thick copper rod passing from top of the
Examples, H2O, N2O, HCl, NH3 Examples, H2, O2 and CO2.
building to the ground. The upper end of the rod has a sharp spike or a
sharp needle. The lower end of the rod is connected to the copper plate
which is buried deep into the ground.

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 1. ELECTROSTATICS 12 PHYSICS EM RAY 8
3. Write a short note on Microwave Oven.
• It is a device which is used to cook and heat the food very quickly. 8. Obtain an expression for the Electric Field due to two parallel
֍ Principle charged infinite sheets.
Torque acting on the electric dipole.
֍ Working
When we switch on the oven it produces oscillating electromagnetic
fields and produce torque on the water molecules present in food. Due to
this torque on each water molecule, it begins to rotate very fast and
produce thermal energy. Thus, heat generated is used to heat the food.

4. During lightning it is safer to sit inside bus than in open ground or

under tree. Why?
During lightning accompanied by a thunderstorm, it is always safer to
sit inside a bus than in open ground or under a tree. The metal body of
the bus provides electrostatic shielding, since the electric field inside is
zero. During lightning, the charges flow through the body of the
Consider two infinitely large charged plane sheets with equal and
conductor to the ground with no effect on the person inside that bus.
opposite charge densities +𝜎 and –𝜎 which are placed parallel to each
other as shown in the following figure.
5. Why the electric field is zero, everywhere inside the conductor?
This is an experimental fact. Suppose the electric field is not zero
֍ The electric field between the plates and outside the plates is found
inside the metal, then there will be a force on the mobile charge carriers
using Gauss law.
due to this electric field. As a result, there will be a net motion of the
• The magnitude of the electric field due to an infinite charged plane
mobile charges, which contradicts the conductors being in electrostatic 𝜎
equilibrium. sheet is 𝐸 = 2𝜀
0
The direction of electric field is perpendicularly outward if 𝜎 > 0.
6. Define point dipole. The direction of electric field is perpendicularly in ward if 𝜎 < 0.
The general expression for the electric dipole is 𝑝 = 2𝑎𝑞, in this At the points P2 and P3, the electric field due to both plates are equal
expression if ‘2a’ approaches to zero and ‘q’ approaches to infinity such in magnitude and opposite in direction. As a result, electric field at a
that the product of 𝑝 = 2𝑎𝑞 is finite, then the dipole is called a point point outside the plates is zero.
dipole. But, inside the plate, electric fields are in same direction. i.e., towards
the right, the total electric field at a point P1, is
7. Write down the properties of electric field lines
• The electric field lines start from a positive charge and end at negative charges or at infinity. 𝝈
𝑬𝒊𝒏𝒔𝒊𝒅𝒆 = (𝟐𝜺 +
𝝈 𝝈
)=𝜺
𝟎 𝟐𝜺𝟎 𝟎
• The electric field lines are never intersect with each other.
• The electric field lines are denser in a region where the electric field and Eoutside = 0

has larger magnitude and less dense in a region where the electric field
is of smaller magnitude.
• The electric field vector at a point in space is tangential to the electric
field line at that point.
• For a positive point charge the electric field lines point radially outward
and for a negative point charge, the electric field lines points radially
inward.
• The number of electric field lines that emanate from the positive charge
or end at a negative charge is directly proportional to the magnitude of
the charges. (No. field lines ∝ Q )
• The magnitude of the electric field for a point charge decreases as the
distance increases.
1
(|𝐸⃗ | ∝ )
𝑟2
GHSS KALAMARUDUR - KALLAKURICHI DISTRICT