An electric dipole is formed by two charges $+q$ and $-q$ located in xy-plane at $(0,2)$ mm and $(0,-2)$ mm, respectively, as shown in the figure. The electric potential at point $\mathrm{P}(100,100)$ mm due to the dipole is $V_{0}$. The charges $+q$ and $-q$ are then moved to the points $(-1,2)$ mm and $(1,-2)$ mm, respectively. What is the value of electric potential at P due to the new dipole ?

(C) $\mathrm{V}_{\mathrm{o}} / \sqrt{2}$

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Electrostatics

An electric dipole is formed by two charges

+ q

and

- q

located in xy-plane at

(0, 2)

mm and

(0, - 2)

mm, respectively, as shown in the figure. The electric potential at point

P (100, 100)

mm due to the dipole is

V_{0}

. The charges

+ q

and

- q

are then moved to the points

(- 1, 2)

mm and

(1, - 2)

mm, respectively. What is the value of electric potential at P due to the new dipole ?

Answer: (B)

The electric potential due to a dipole at a point is given by

V = \frac{K P \cdot r}{r ^{3}}

.

Initially,

P_{1} = 4 q \hat{j}

(dipole moment magnitude

4 q

mm along y-axis) and

r = 100 (\hat{i} + \hat{j})

mm. This gives

V_{0} = \frac{K ( 100 \cdot 4 q )}{( 100 2 ) ^{3}}

.

After moving the charges, the new dipole has moment

P_{2} = q ℓ

where

ℓ = 22

mm at angle

θ

with

tan θ = 2

. The dipole moment components are

P_{2} = P_{2} [- cos θ \hat{i} + sin θ \hat{j}]

.

The new potential is

V = \frac{K P _{2} \cdot r}{r ^{3}} = \frac{K ( 100 P _{2} ) ( - c o s θ + s i n θ )}{( 100 2 ) ^{3}}

.

With

cos θ = 1/ 5

and

sin θ = 2/ 5

, we get

- cos θ + sin θ = 1/ 5

.

Since

P_{2} = q \cdot 22 \cdot 5 / 5 = 2 q 2

, the ratio gives:

V = V_{0} \cdot \frac{2 q 2}{4 q} \cdot (- 1/ 5 + 2/ 5) \cdot 5 = V_{0} \cdot \frac{1}{2} \cdot 1 = \frac{V _{0}}{2}

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Single correct option