List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude \(p\), oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance \(2 r\) apart along the \(x\) direction. The midpoint of the line joining the two dipoles is \(X\). The possible resultant electric fields \(\vec{E}\) at \(X\) 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-I (P) (see structure) (Q) (see structure) (R) (see structure) (S) (see structure) List-II (1) \(\vec{E}=0\) (2) \(\vec{E}=-\frac{p}{2 \pi \epsilon_{0} r^{3}} \hat{\mathrm{j}}\) (3) \(\vec{E}=-\frac{p}{4 \pi \epsilon_{0} r^{3}}(\hat{\mathrm{i}}-\hat{\mathrm{j}})\) (4) \(\vec{E}=\frac{p}{4 \pi \epsilon_{0} r^{3}}(2 \hat{i}-\hat{j})\) (5) \(\vec{E}=\frac{p}{\pi \epsilon_{0} \mathrm{r}^{3}} \hat{\mathrm{i}}\)

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Electrostatics

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude

p

, oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance

2 r

apart along the

x

direction. The midpoint of the line joining the two dipoles is

X

. The possible resultant electric fields

E

X

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-I

(P) (see structure)
(Q) (see structure)
(R) (see structure)
(S) (see structure)

List-II

(1)

E = 0

(2)

E = - \frac{p}{2 π ϵ _{0} r ^{3}} \hat{j}

(3)

E = - \frac{p}{4 π ϵ _{0} r ^{3}} (\hat{i} - \hat{j})

(4)

E = \frac{p}{4 π ϵ _{0} r ^{3}} (2 \hat{i} - \hat{j})

(5)

E = \frac{p}{π ϵ _{0} r ^{3}} \hat{i}

(A)P→2, Q→1, R→3, S→5
(B)P→4, Q→5, R→3, S→1
(C)P→2, Q→1, R→4, S→5✓ Correct
(D)P→2, Q→1, R→3, S→5

Answer: (C)

Solution

For each configuration, we calculate the electric field at the midpoint X due to both dipoles:

(P) Both dipoles vertical, pointing up: Each contributes

\frac{- k p}{r ^{3}} \hat{j}

at the midpoint (axial position). Total:

E = \frac{- 2 k p}{r ^{3}} \hat{j} = \frac{- p}{2 π ϵ _{0} r ^{3}} \hat{j}

→ (2)

(Q) Dipoles pointing in opposite directions along x: Fields cancel.

E = 0

→ (1)

(R) Both dipoles at 45° to x-axis: Each contributes components in both

\hat{i}

and

\hat{j}

directions. Net field:

E = \frac{p}{4 π ϵ _{0} r ^{3}} (2 \hat{i} - \hat{j})

→ (4)

(S) Both dipoles horizontal, pointing right: Each contributes

\frac{2 k p}{r ^{3}} \hat{i}

at the midpoint (equatorial position). Total:

E = \frac{4 k p}{r ^{3}} \hat{i} = \frac{p}{π ϵ _{0} r ^{3}} \hat{i}

→ (5)

Therefore: P→2, Q→1, R→4, S→5

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option

Practise more: all JEE Advanced PYQs on Electrostatics →

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Electrostatics

Solution

More JEE Advanced questions on Electrostatics