JEE Advanced 2023 — Paper 2 — PHY-I solution

JEE Advanced2023Paper 2PHY-I

An electric dipole is formed by two charges $+ q$ and $- q$ located in $x y$ -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?

A
$V_{0} /4$
B
$V_{0} /2$
C
$V_{0} / 2$
D
$3 V_{0} /4$

Solution

Original dipole: The dipole moment has magnitude $p_{1} = q (4 mm)$ and points along $\hat{j}$ , so $p_{1} = 4 q \hat{j}$ (in mm units).

The position vector from the dipole centre (origin) to $P$ is $r = 100 (\hat{i} + \hat{j})$ mm with magnitude $r = 1002$ mm. The potential of a point dipole is

$V = \frac{k p \cdot r}{r ^{3}}$ .

Hence $V_{0} = \frac{k ( 4 q ) ( 100 )}{( 100 2 ) ^{3}}$ .

New dipole: The vector from $- q$ at $(1, - 2)$ to $+ q$ at $(- 1, 2)$ is $(- 2, 4)$ mm, so $p_{2} = q (- 2 \hat{i} + 4 \hat{j})$ mm. Taking the dot product with $r = 100 (\hat{i} + \hat{j})$ gives $p_{2} \cdot r = 100 q (- 2 + 4) = 200 q$ .

Therefore $V_{new} = \frac{k ( 200 q )}{( 100 2 ) ^{3}}$ .

Taking the ratio, $\frac{V _{new}}{V _{0}} = \frac{200 q}{400 q} = \frac{1}{2}$ .

$∴ V_{new} = \frac{V _{0}}{2}$ .

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