An electric dipole is formed by two charges and located in the -plane at (0, 2) mm and mm, respectively, as shown in the figure. The electric potential at point due to the dipole is . The charges and are then moved to the points and , respectively.

Question

An electric dipole is formed by two charges $+ q$ and $- q$ located in the $x y$ -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 [2023]

Smart Achievers · Accepted Answer

(2)

Since electric potential due to a dipole at a general point is given by

$V = \frac{K p \cos θ}{r^{3}}$

$V_{1} \propto \frac{p_{1} \cos θ_{1}}{r_{1}^{3}}$

Electric potential at point $P (100, 100)$ $V_{1} = V_{0}$ , and the electric potential at the new point is

$V_{2} \propto \frac{p_{2} \cos θ_{2}}{r_{2}^{3}}$

$\frac{V_{2}}{V_{1}} = \frac{p_{2} \cos θ_{2}}{p_{1} \cos θ_{1}} = \frac{q (- 2 \hat{i} + 4 \hat{j}) \cdot (\hat{i} + \hat{j})}{q (0 \hat{i} + 4 \hat{j}) \cdot (\hat{i} + \hat{j})} = \frac{1}{2}$

$∴$ $V_{new} = \frac{V_{0}}{2}$

Solution

1 $V_{0} / 4$

2 $V_{0} / 2$

3 $V_{0} / \sqrt{2}$

4 $3 V_{0} / 4$

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Solution

1 V0/4

2 V0/2

3 V0/2

4 3V0/4

Ans. (2) Since electric potential due to a dipole at a general point is given by V=Kpcosθr3 V1∝p1cosθ1r13 Electric potential at point P(100,100)V1=V0, and the electric potential at the new point is V2∝p2cosθ2r23 V2V1=p2cosθ2p1cosθ1=q(-2i^+4j^)·(i^+j^)q(0i^+4j^)·(i^+j^)=12 ∴ Vnew=V02

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1 $V_{0} / 4$

2 $V_{0} / 2$

3 $V_{0} / \sqrt{2}$

4 $3 V_{0} / 4$