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(A) For an uncharged conducting sphere,the net charge is zero. When placed in an external electric field (due to the dipole),the induced charges on th

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$\frac{kp \cos \phi}{r^2}$

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(A) For an uncharged conducting sphere,the net charge is zero. When placed in an external electric field (due to the dipole),the induced charges on the surface redistribute such that the potential at every point on the surface of the conductor is the same and equal to the potential at the center of the sphere.The potential at the center of the sphere due to the dipole is $V_{	ext{center}} = \frac{kp \cos 	heta}{d^2}$,where $d$ is the distance from the dipole to the center and $	heta$ is the angle between the dipole moment and the position vector of the center.However,based on the geometry provided in the diagram,the potential at point $A$ is the sum of the potential due to the dipole and the potential due to the induced charges on the sphere. Since the sphere is uncharged and isolated,the net potential due to induced charges at the center is zero. Thus,the potential at the surface is equal to the potential due to the dipole at the center of the sphere. Given the geometry where the distance from the dipole to the center is $d = r \cos \phi$,the potential is $V = \frac{kp \cos \phi}{d^2} = \frac{kp \cos \phi}{(r \cos \phi)^2} = \frac{kp \cos \phi}{r^2 \cos^2 \phi} = \frac{kp}{r^2 \cos \phi}$.Wait,re-evaluating the standard result for this specific configuration: The potential at the surface of an uncharged conducting sphere in the presence of a dipole is $V = \frac{kp \cos \phi}{r^2}$ where $r$ is the distance from the dipole to the point $A$.

$A$ dipole having dipole moment $p$ is placed in front of a solid uncharged conducting sphere as shown in the diagram. The net potential at point $A$ lying on the surface of the sphere is :-

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