A dipole of dipole moment p is placed at a distance 2R from the centre of a neutral conducting sphere of radius R. The direction of dipole is towards the centre of the sphere. A tangent is drawn from the dipole to the sphere which meets the sphere at point A. Mark the incorrect option.

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The potential at point A is $p / 16 π ε_{0} R^{2}$

The potential at point A is $\sqrt{3} p / 48 π ε_{0} R^{2}$

The potential at point A due to induced charges is $(3 - 2 \sqrt{3}) p / 48 π ε_{0} R^{2}$

The potential at center of sphere is $p / 16 π ε_{0} R^{2}$

answer is B.

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Detailed Solution

Potential at center and at any point on the sphere will be same

So, $V_{A} = V_{C} = \frac{P}{4 π \in_{0} {(2 R)}^{2}} = \frac{P}{16 π \in_{0} R^{2}}$

${(V_{A})}_{d i p o l e} = \frac{P \cos 30^{o}}{4 π \in_{0} {(R \sqrt{3})}^{2}}$

${(V_{A})}_{D i p o l e} = \frac{\sqrt{3} P}{24 π \in_{0} R^{2}}$

${(V_{A})}_{I n} = V_{A} - {(V_{A})}_{D i p o l e}$

${(V_{A})}_{I n} = \frac{P (3 - 2 \sqrt{3})}{48 π \in_{0} R^{2}}$

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