Question 14

The electric field of an ideal dipole at distance $r$ is given by:

• Axial field: $\vec{E}_{ax} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3}$ (same direction as $\vec{p}$)
• Equatorial field: $\vec{E}_{eq} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3}$ (opposite direction to $\vec{p}$)

Let $k = \frac{1}{4\pi\epsilon_0 r^3}$.

(P) Both dipoles are $\vec{p} = p\hat{j}$. Point $X(0,0)$ is on the equatorial line for both dipoles located at $x = \pm r$. $\vec{E}_X = \vec{E}_L + \vec{E}_R = -kp\hat{j} - kp\hat{j} = -2kp\hat{j} = -\frac{p}{2\pi\epsilon_0 r^3} \hat{j}$. Matches (2).

(Q) Left dipole is $p\hat{j}$, right is $-p\hat{j}$. Point $X$ is on the equatorial line for both. $\vec{E}_X = -kp\hat{j} - k(-p\hat{j}) = -kp\hat{j} + kp\hat{j} = 0$. Matches (1).

(R) Left dipole is $p\hat{j}$ (equatorial for $X$), right is $p\hat{i}$ (axial for $X$). $\vec{E}_L = -kp\hat{j}$ $\vec{E}_R = 2kp\hat{i}$ Total $\vec{E} = kp(2\hat{i} - \hat{j}) = \frac{p}{4\pi\epsilon_0 r^3} (2\hat{i} - \hat{j})$. Matches (4).

(S) Both dipoles are $p\hat{i}$. Both are axial to $X$. $\vec{E}_L = 2kp\hat{i}$ $\vec{E}_R = 2kp\hat{i}$ Total $\vec{E} = 4kp\hat{i} = \frac{4p}{4\pi\epsilon_0 r^3} \hat{i} = \frac{p}{\pi\epsilon_0 r^3} \hat{i}$. Matches (5).

Correct matching: P-2, Q-1, R-4, S-5. This corresponds to option (C).