Electric Field and Charges · Case 1: If we wish to calculate the electric field intensity due to a uniformly charged spherical shell at a point outside the shell i.e. N> . Let point

Electric Field and

ChargesModule 5 of 5

Class: XII

Subject: Physics

Month: April

Contents

Electric Field Intensity due to a Uniformly Charged Spherical

Shell

Electric Field due to an Electric Dipole at an Equatorial Point

Torque on an Electric Dipole placed in a Uniform Electric Field

6. Electric Field Intensity due to a Uniformly Charged Spherical

Shell

In the case of a uniformly charged spherical shell, we consider the

surface charge density of the spherical shell.

The surface charge density is defined as the total charge present

per unit surface area of the spherical shell. It is denoted by σ.

Here, we assume a positive charge q to be distributed on the

surface of a spherical shell of radius R.

𝜎 =𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑠ℎ𝑒𝑙𝑙

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑖𝑐𝑎𝑙 𝑠ℎ𝑒𝑙𝑙=𝑞

𝐴

Therefore,

𝑞 = 𝜎𝐴 = 𝜎 × 4𝜋𝑅2

Application of Gauss’ Law for Special

Cases

Case 1: If we wish to calculate the electric field intensity due to a

uniformly charged spherical shell at a point outside the shell i.e.

𝑟 > 𝑅.

Let point P be situated outside the spherical shell at a distance r

from the center of the shell. Thus, 𝑟 > 𝑅.

In order to calculate the electric field intensity at a distance r

from the spherical shell, we assume a spherical Gaussian surface

with radius r and concentric with the spherical shell.


Cases

Hence, the total electric flux over the spherical Gaussian surface

of radius r will be,

ϕ𝐸 = ර𝐸 ∙ 𝑑𝑆 = ර𝐸𝑑𝑆 cos 𝜃

Due to the spherical geometry of the Gaussian surface, 𝐸 and 𝑑𝑆will be parallel to each other at each and every point on the

Gaussian surface. Hence, the angle between them will be 0o.

ϕ𝐸 = ර𝐸𝑑𝑆 cos 0 = ර𝐸𝑑𝑆 = 𝐸ර𝑑𝑆

𝑑𝑆ׯ is the total surface area of the spherical Gaussian surface of

radius r which is equal to 4π𝑟2.

ϕ𝐸 = 𝐸 × 4π𝑟2 (1)


Cases

But according to Gauss’ law,

ϕ𝐸 = 𝑆ׯ 𝐸 . 𝑑𝑆 =1

∈0× 𝑞

where, q is the total charge enclosed by the spherical Gaussian

surface.

ϕ𝐸 =1

∈0× 𝑞 (2)

From equations (1) and (2) we get,

𝐸 × 4π𝑟2 =1

∈0× 𝑞

𝐸 =𝑞

4𝜋𝑟2 ∈0


Cases

On substituting 𝑞 = σ × 4π𝑅2 we get,

𝐸 =σ × 4π𝑅2

4𝜋𝑟2 ∈0

𝐸 =σ𝑅2

𝑟2 ∈0

Thus, the electric field intensity due to a uniformly charged

spherical shell at a distance r from its center is inversely

proportional to the square of distance from it.

The electric charge distribution over the surface of a spherical

shell behaves as if the whole charge were situated at the center

of the spherical shell.


Cases


uniformly charged spherical shell at its surface i.e. 𝑟 = 𝑅.

In this case, the point P lies on the surface of the spherical shell.

Hence, we substitute 𝑟 = 𝑅 in the equation above.

𝐸 =σ𝑅2

𝑅2 ∈0

𝐸 =σ

∈0


uniformly charged spherical shell at a point inside the shell i.e.

𝑟 < 𝑅.

Let point P be situated inside the spherical shell at a distance r

from the center of the shell. Thus, 𝑟 < 𝑅.


Cases

In order to calculate the electric field intensity at point P which is

at a distance r from the center of the spherical shell, we assume a

spherical Gaussian surface with radius r and concentric with the

spherical shell.

Hence, the total electric flux over the spherical Gaussian surface

of radius r will be,

ϕ𝐸 = ර𝐸 ∙ 𝑑𝑆 = ර𝐸𝑑𝑆 cos 𝜃


Cases

Due to the spherical geometry of the Gaussian surface, 𝐸 and 𝑑𝑆will be parallel to each other at each and every point on the

Gaussian surface. Hence, the angle between them will be 0o.

ϕ𝐸 = ර𝐸𝑑𝑆 cos 0 = ර𝐸𝑑𝑆 = 𝐸ර𝑑𝑆

𝑑𝑆ׯ is the total surface area of the spherical Gaussian surface of

radius r which is equal to 4π𝑟2.

ϕ𝐸 = 𝐸 × 4π𝑟2 (1)

But according to Gauss’ law,

ϕ𝐸 = 𝑆ׯ 𝐸 . 𝑑𝑆 =1

∈0× 𝑞

where, q is the total charge enclosed by the spherical Gaussian

surface.


Cases

ϕ𝐸 =1

∈0× 𝑞 (2)

Since the charge q is distributed on the surface of the spherical

shell, there will be no charge enclosed by the spherical Gaussian

surface i.e. 𝑞 = 0.

Hence,

ϕ𝐸 =1

∈0× 𝑞 =

1

∈0× 0 = 0

From equations (1) and (2) we get,

𝐸 × 4π𝑟2 = 0

Therefore,

𝐸 = 0

Hence, there is no electric field inside a uniformly charged

spherical shell.


Cases

In the figure shown above, point O is called the midpoint of the

electric dipole and point P is called the equatorial point of the

dipole. Distance of the equatorial point from the midpoint is r.

Electric Field due to an Electric Dipole

at an Equatorial Point

Therefore, the equatorial distance 𝑂𝑃 = 𝑟.

Since point P is in the neighbourhood of the two charges which

are equal in magnitude and opposite in nature, there exists two

kinds of electric field intensities at point P.

Electrical field intensity at point P due to charge +𝑞 is,

𝐸1 = 𝑘𝑞

𝑟2 + 𝑎22= 𝑘

𝑞

𝑟2 + 𝑎2

having direction away from +𝑞 charge.

Electrical field intensity at point P due to charge −𝑞 is,

𝐸2 = 𝑘𝑞

𝑟2 + 𝑎22= 𝑘

𝑞

𝑟2 + 𝑎2

having direction towards −𝑞 charge.



Let the angle between 𝐸1 and 𝐸2 be 2𝜃.

Using Parallelogram Law of Vector Addition, the resultant

electric field intensity at point P is given as,

𝐸 = 𝐸12 + 𝐸2

2 + 2𝐸1𝐸2 cos 2𝜃

We know that,

𝐸1 = 𝐸2 = 𝑘𝑞

𝑟2 + 𝑎2

i.e. both the electric field intensities have the same magnitude.

Thus, by using 𝐸1 = 𝐸2 we get,

𝐸 = 𝐸12 + 𝐸1

2 + 2𝐸1𝐸1 cos 2𝜃

𝐸 = 2𝐸12 + 2𝐸1

2 cos 2𝜃



𝐸 = 2𝐸12 1 + cos 2𝜃

Using the trigonometric identity cos 2𝜃 = 2𝑐𝑜𝑠2𝜃 − 1, we have

1 + cos 2𝜃 = 2𝑐𝑜𝑠2𝜃

Thus,

𝐸 = 2𝐸12 2𝑐𝑜𝑠2𝜃 = 4𝐸1

2𝑐𝑜𝑠2𝜃

𝐸 = 2𝐸1 cos 𝜃

From the figure given above, it can be seen that,

cos 𝜃 =𝑎

𝑟2 + 𝑎2

Therefore,

𝐸 = 2𝐸1𝑎

𝑟2 + 𝑎2



Using 𝐸1 = 𝑘𝑞

𝑟2+𝑎2we get,

𝐸 = 2𝑘𝑞

𝑟2 + 𝑎2𝑎

𝑟2 + 𝑎2

𝐸 = 𝑘𝑞 × 2𝑎

𝑟2 + 𝑎2 ൗ3 2

Since 𝑝 = 𝑞(2𝑎) we have,

𝐸 =𝑘𝑝

𝑟2 + 𝑎2 ൗ3 2

If the equatorial distance of the point considered here is much

larger than the length of the dipole i.e. 𝑟 ≫ 𝑎 then 𝑟2 ≫ 𝑎2.



Hence, we can neglect 𝑎2 and the equation reduces to

𝐸 =𝑘𝑝

𝑟2 ൗ3 2

𝐸 =𝑘𝑝

𝑟3

Therefore, it can be concluded that the electrical field

intensity at an equatorial point of an electric dipole is always

antiparallel to the electric dipole moment of that electric

dipole and the angle between these two vectors is 180o or π

rad.



Consider an electric dipole placed in a uniform electric field

such that it makes an angle θ with the electric field.

Since the electric dipole consists of two equal and opposite

charges 𝑞 and −𝑞, each of the two charges will experience a

force when placed in a uniform electric field.

Torque on an Electric Dipole placed in

a Uniform Electric Field

Force experienced by charge 𝑞 when placed in a uniform

electric field E is given as,

𝐹1 = 𝑞𝐸

This force will be directed parallel to the electric field since it is a

positive charge.

Force experienced by charge −𝑞 when placed in a uniform

electric field E is given as,

𝐹2 = −𝑞𝐸

This force will be directed antiparallel to the electric field since it

is a negative charge.

Therefore, the resultant force on the electric dipole is given as,

𝐹 = 𝐹1 + 𝐹2𝐹 = 𝑞𝐸 + −𝑞𝐸 = 𝑞𝐸 − 𝑞𝐸 = 0



Hence, the net force on an electric dipole placed in a uniform

electric field is zero.

However, the line of action of these two forces is not the same.

Since the forces are equal in magnitude, opposite in direction

and have a line of action that does not coincide, they form a

couple.

Due to this couple, the electric dipole experiences a torque

that tends to rotate it in the direction of electric field.

This torque is given as the cross product of either of the forces

and the perpendicular distance between the two forces.

𝜏 = (𝑞𝐸) × (2𝑙 sin 𝜃)

On rearranging this equation, we get,

𝜏 = (𝑞 × 2𝑙)𝐸 sin 𝜃



We know that the electric dipole moment 𝑝 = 𝑞(2𝑙)𝜏 = 𝑝𝐸 sin 𝜃

In vector notation this can be written as,

Ԧ𝜏 = Ԧ𝑝 × 𝐸

Thus, the torque acting on an electric dipole may or may not be

zero since it depends on the angle between Ԧ𝑝 and 𝐸.

Hence, we conclude that the net force on an electric dipole

placed in a uniform electric field is zero i.e. there is no

translatory force on the dipole in the electric field but the

net torque on an electric dipole placed in a uniform electric

field may or may not be zero depending upon the orientation of

the dipole in the electric field i.e. there may or may not be

rotation of the dipole in the electric field.



Documents

Electric Field and Charges · Case 1: If we wish to calculate the electric field intensity due to a uniformly charged spherical shell at a point outside the shell i.e. N> . Let point