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Electrostatics question

2025 · Shift 1 · Q46
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Electrostatics question

2025 · Shift 1 · Q46

JEE AdvancedPhysicsElectrostaticsMCQ+4 / −1

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude ppp, oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance 2r2 r2r apart along the xxx direction. The midpoint of the line joining the two dipoles is XXX. The possible resultant electric fields E⃗\vec{E}E at XXX 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 List–II
(P) JEE Advanced 2025 Paper 1 Online Physics - Electrostatics Question 1 English 1 (1) E⃗=0\vec{E}=0E=0
(Q) JEE Advanced 2025 Paper 1 Online Physics - Electrostatics Question 1 English 2 (2) E⃗=− p2πϵ0 r3 ȷ^\displaystyle \vec{E} = -\,\frac{p}{2\pi\epsilon_0\,r^3}\,\hat{\jmath}E=−2πϵ0​r3p​^​
(R) JEE Advanced 2025 Paper 1 Online Physics - Electrostatics Question 1 English 3 (3) E⃗=− p4πϵ0 r3 (ı^−ȷ^)\displaystyle \vec{E} = -\,\frac{p}{4\pi\epsilon_0\,r^3}\,(\hat{\imath} - \hat{\jmath})E=−4πϵ0​r3p​(^−^​)
(S) JEE Advanced 2025 Paper 1 Online Physics - Electrostatics Question 1 English 4 (4) E⃗=p4πϵ0 r3 (2ı^−ȷ^)\displaystyle \vec{E} = \frac{p}{4\pi\epsilon_0\,r^3}\,(2\hat{\imath} - \hat{\jmath})E=4πϵ0​r3p​(2^−^​)
(5) E⃗=pπϵ0 r3 ı^\displaystyle \vec{E} = \frac{p}{\pi\epsilon_0\,r^3}\,\hat{\imath}E=πϵ0​r3p​^
  1. A
    P→3,Q→1,R→2, S→4\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 4P→3,Q→1,R→2, S→4
  2. B
    P→4,Q→5,R→3, S→1\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 1P→4,Q→5,R→3, S→1
  3. C
    P→2,Q→1,R→4, S→5\mathrm{P} \rightarrow 2, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 5P→2,Q→1,R→4, S→5
  4. D
    P→2,Q→1,R→3, S→5\mathrm{P} \rightarrow 2, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 5P→2,Q→1,R→3, S→5
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