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PhyMedAdv 2025 • Paper 1

Match Dipole Configurations with Resultant Electric Field at Midpoint

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 2r2r 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.

Question Diagram 1

Options

A

P→\rightarrow3, Q→\rightarrow1, R→\rightarrow2, S→\rightarrow4

B

P→\rightarrow4, Q→\rightarrow5, R→\rightarrow3, S→\rightarrow1

C

P→\rightarrow2, Q→\rightarrow1, R→\rightarrow4, S→\rightarrow5

Correct
D

P→\rightarrow2, Q→\rightarrow1, R→\rightarrow3, S→\rightarrow5

Topics & Concepts

Electric Charges and FieldsElectric FieldCoulomb's Law

Step-by-Step Solution

To find the correct matching between the dipole configurations in List-I and the resultant electric fields at midpoint XX in List-II, we use the general formula for the electric field due to an ideal electric dipole with dipole moment p⃗\vec{p} at a position vector r⃗\vec{r} relative to the dipole:

E⃗(r⃗)=14πϵ0r3[3(p⃗⋅r^)r^−p⃗]\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0 r^3} \left[ 3(\vec{p} \cdot \hat{r})\hat{r} - \vec{p} \right]

Let point XX be placed at the origin (0,0,0)(0,0,0).

  • The left dipole is located at position vector −ri^-r\hat{i}, so the displacement vector from the left dipole to XX is r⃗1=ri^\vec{r}_1 = r\hat{i} with unit vector r^1=i^\hat{r}_1 = \hat{i}.
  • The right dipole is located at position vector +ri^+r\hat{i}, so the displacement vector from the right dipole to XX is r⃗2=−ri^\vec{r}_2 = -r\hat{i} with unit vector r^2=−i^\hat{r}_2 = -\hat{i}.

Configuration (P)

  • Left dipole: p⃗1=pj^\vec{p}_1 = p\hat{j} E⃗1=14πϵ0r3[3(pj^⋅i^)i^−pj^]=−pj^4πϵ0r3\vec{E}_1 = \frac{1}{4\pi\epsilon_0 r^3}\left[ 3(p\hat{j} \cdot \hat{i})\hat{i} - p\hat{j} \right] = -\frac{p\hat{j}}{4\pi\epsilon_0 r^3}

  • Right dipole: p⃗2=pj^\vec{p}_2 = p\hat{j} E⃗2=14πϵ0r3[3(pj^⋅(−i^))(−i^)−pj^]=−pj^4πϵ0r3\vec{E}_2 = \frac{1}{4\pi\epsilon_0 r^3}\left[ 3(p\hat{j} \cdot (-\hat{i}))(-\hat{i}) - p\hat{j} \right] = -\frac{p\hat{j}}{4\pi\epsilon_0 r^3}

  • Resultant Electric Field at XX: E⃗net=E⃗1+E⃗2=−2pj^4πϵ0r3=−p2πϵ0r3j^\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = -\frac{2p\hat{j}}{4\pi\epsilon_0 r^3} = -\frac{p}{2\pi\epsilon_0 r^3} \hat{j}

Thus, P→2\text{P} \rightarrow 2.

Configuration (Q)

  • Left dipole: p⃗1=pj^\vec{p}_1 = p\hat{j} E⃗1=−pj^4πϵ0r3\vec{E}_1 = -\frac{p\hat{j}}{4\pi\epsilon_0 r^3}

  • Right dipole: p⃗2=−pj^\vec{p}_2 = -p\hat{j} E⃗2=14πϵ0r3[3(−pj^⋅(−i^))(−i^)−(−pj^)]=pj^4πϵ0r3\vec{E}_2 = \frac{1}{4\pi\epsilon_0 r^3}\left[ 3(-p\hat{j} \cdot (-\hat{i}))(-\hat{i}) - (-p\hat{j}) \right] = \frac{p\hat{j}}{4\pi\epsilon_0 r^3}

  • Resultant Electric Field at XX: E⃗net=E⃗1+E⃗2=0\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = 0

Thus, Q→1\text{Q} \rightarrow 1.

Configuration (R)

  • Left dipole: p⃗1=pj^\vec{p}_1 = p\hat{j} E⃗1=−pj^4πϵ0r3\vec{E}_1 = -\frac{p\hat{j}}{4\pi\epsilon_0 r^3}

  • Right dipole: p⃗2=pi^\vec{p}_2 = p\hat{i} E⃗2=14πϵ0r3[3(pi^⋅(−i^))(−i^)−pi^]=14πϵ0r3[3(−p)(−i^)−pi^]=2pi^4πϵ0r3\vec{E}_2 = \frac{1}{4\pi\epsilon_0 r^3}\left[ 3(p\hat{i} \cdot (-\hat{i}))(-\hat{i}) - p\hat{i} \right] = \frac{1}{4\pi\epsilon_0 r^3}\left[ 3(-p)(-\hat{i}) - p\hat{i} \right] = \frac{2p\hat{i}}{4\pi\epsilon_0 r^3}

  • Resultant Electric Field at XX: E⃗net=E⃗1+E⃗2=p4πϵ0r3(2i^−j^)\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = \frac{p}{4\pi\epsilon_0 r^3} (2\hat{i} - \hat{j})

Thus, R→4\text{R} \rightarrow 4.

Configuration (S)

  • Left dipole: p⃗1=pi^\vec{p}_1 = p\hat{i} E⃗1=14πϵ0r3[3(pi^⋅i^)i^−pi^]=2pi^4πϵ0r3\vec{E}_1 = \frac{1}{4\pi\epsilon_0 r^3}\left[ 3(p\hat{i} \cdot \hat{i})\hat{i} - p\hat{i} \right] = \frac{2p\hat{i}}{4\pi\epsilon_0 r^3}

  • Right dipole: p⃗2=pi^\vec{p}_2 = p\hat{i} E⃗2=2pi^4πϵ0r3\vec{E}_2 = \frac{2p\hat{i}}{4\pi\epsilon_0 r^3}

  • Resultant Electric Field at XX: E⃗net=E⃗1+E⃗2=4pi^4πϵ0r3=pπϵ0r3i^\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 = \frac{4p\hat{i}}{4\pi\epsilon_0 r^3} = \frac{p}{\pi\epsilon_0 r^3} \hat{i}

Thus, S→5\text{S} \rightarrow 5.

Conclusion

The correct match is: P→2,Q→1,R→4,S→5\text{P} \rightarrow 2, \quad \text{Q} \rightarrow 1, \quad \text{R} \rightarrow 4, \quad \text{S} \rightarrow 5

This matches Option C.

Match Dipole Configurations with Resultant Electric Field at Midpoint | Physics PYQ Solution - JEE Challenger