Two electric dipoles $\mathrm{A}, \mathrm{B}$ with respective dipole moments $\overrightarrow{\mathrm{d}_{\mathrm{A}}}=-4$ qai and $\overrightarrow{\mathrm{d}_{\mathrm{B}}}=2$ qai are placed on the $x$-axis with a separation R , as shown in the figure. The distance from A at which both of them produce the same potential is:

(B) $\frac{\sqrt{2} \mathrm{R}}{\sqrt{2}+1}$

(A) $\frac{\mathrm{R}}{\sqrt{2}+1}$

(C) $\frac{\mathrm{R}}{\sqrt{2}-1}$

(D) $\frac{\sqrt{2} R}{\sqrt{2}-1}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Electrostatics

Two electric dipoles

A, B

with respective dipole moments

d_{A} = - 4

qai and

d_{B} = 2

qai are placed on the

x

-axis with a separation R , as shown in the figure. The distance from A at which both of them produce the same potential is:

Answer: (B)

Place

A

at the origin and

B

x = R

with both dipole moments along the axis. On a dipole's axis the potential has magnitude

\frac{k p}{r ^{2}}

, so equating the two contributions at a point a distance

x

from

A

and

R - x

from

B

gives

\frac{k ( 4 q a )}{x ^{2}} = \frac{k ( 2 q a )}{( R - x ) ^{2}}

. Then

2 (R - x)^{2} = x^{2}

, i.e.

2 (R - x) = x

, which solves to

x = \frac{2 R}{2 + 1}

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option