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Physics · Electrostatics

JEE Advanced 2025 — Paper 1 — Question 12

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude pp, oriented as marked by arrows in the figures.

In all the configurations the dipoles are fixed such that they are at a distance 2r2 r apart along the xx direction. The midpoint of the line joining the two dipoles is XX. The possible resultant electric fields E⃗\vec{E} at XX are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II.

List-IList-II
(P) figure(1) E⃗=0\vec{E}=0
(Q) figure(2) E⃗=−p2πϵ0r3j \vec{E}=-\frac{p}{2\pi {{\epsilon }_{0}}{{\text{r}}^{3}}}\overset{\text{}}{\mathop{\text{j}}}\,
(R) figure(3) E⃗=−p4πϵ0r3(i −j )\vec{E}=-\frac{p}{4\pi {{\epsilon }_{0}}{{\text{r}}^{3}}}\left( \overset{\text{}}{\mathop{\text{i}}}\,-\overset{\text{}}{\mathop{\text{j}}}\, \right)
(S) figure(4) E⃗=p4πϵ0r3(2i −j )\vec{E}=\frac{p}{4\pi {{\epsilon }_{0}}{{\text{r}}^{3}}}\left( 2\overset{\text{}}{\mathop{\text{i}}}\,-\overset{\text{}}{\mathop{\text{j}}}\, \right)
(5) E⃗=pπϵ0r3i \vec{E}=\frac{p}{\pi {{\epsilon }_{0}}{{r}^{3}}}\overset{\text{}}{\mathop{\text{i}}}\,
  1. Option A:

    ) P→3,Q→1,R→2, S→4\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 4

  2. Option B:

    P→4,Q→5,R→3, S→1\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 1

  3. Option C:

    P→2,Q→1,R→4, S→5\mathrm{P} \rightarrow 2, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 5

    Correct
  4. Option D:

    P→2,Q→1,R→3, S→5\mathrm{P} \rightarrow 2, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 5

Answer: C

Step-by-step solution

(p) figure Enet  ⁣ ⁣  ⁣ ⁣ =2kPr3j {{E}_{\text{net }\!\!~\!\!\text{ }}}=\frac{2kP}{{{r}^{3}}}\overset{}{\mathop{j}}\, Enet  ⁣ ⁣  ⁣ ⁣ =−Pj 2πϵ0r3{{E}_{\text{net }\!\!~\!\!\text{ }}}=\frac{-P\overset{}{\mathop{j}}\,}{2\pi {{\epsilon }_{0}}{{r}^{3}}}
(Q) figure Enet=0{{\text{E}}_{\text{net}}}=0
(R) figure 2Pi 4πϵ0r3−Pj 4πϵ0r3\frac{2P\overset{}{\mathop{i}}\,}{4\pi {{\epsilon }_{0}}{{r}^{3}}}-\frac{P\overset{}{\mathop{j}}\,}{4\pi {{\epsilon }_{0}}{{r}^{3}}}
(S) figure Enet=4kPi r3{{\text{E}}_{\text{net}}}=\frac{4\text{kP}\overset{\text{}}{\mathop{\text{i}}}\,}{{{\text{r}}^{3}}}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2025
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics
Topic
Electric Dipole