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Question

$:$ If we want to find potential at $A$

$V_{A}=\frac{K_{p} \cos \phi}{r^{2}}$

$V_{A}=V_{0} ightarrow$

$V_{0}=V_{	ext {ind.}}+V_{	ext {dip

Vedclass · Accepted Answer

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

Vedclass · Answer

$:$ If we want to find potential at $A$

$V_{A}=\frac{K_{p} \cos \phi}{r^{2}}$

$V_{A}=V_{0} ightarrow$

$V_{0}=V_{	ext {ind.}}+V_{	ext {dipol.;}}$

$V_{i n d=0}$

$V_{0}=\frac{K_{p} \cos ^{2} \phi}{r^{2}}$

$V_{A}=V_{0}=\frac{K_{p} \cos ^{2} \phi}{r^{2}}$

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

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