A thin conducting spherical shell of radius R | vedclass

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Question

(C) The total electric potential $V$ at point $P$ is the sum of the potential due to the charge $q$ on the shell and the potential due to the charge $

Vedclass · Accepted Answer

$\frac{2Q}{4\pi \epsilon_0 R} + \frac{q}{4\pi \epsilon_0 R}$

Vedclass · Answer

(C) The total electric potential $V$ at point $P$ is the sum of the potential due to the charge $q$ on the shell and the potential due to the charge $Q$ at the center.For a point inside a conducting shell,the potential due to the shell is constant and equal to the potential at its surface,which is $V_{shell} = \frac{1}{4\pi \epsilon_0} \frac{q}{R}$.The potential due to the point charge $Q$ at a distance $r = R/2$ is $V_{point} = \frac{1}{4\pi \epsilon_0} \frac{Q}{R/2} = \frac{1}{4\pi \epsilon_0} \frac{2Q}{R}$.Therefore,the total potential $V = V_{shell} + V_{point} = \frac{q}{4\pi \epsilon_0 R} + \frac{2Q}{4\pi \epsilon_0 R} = \frac{1}{4\pi \epsilon_0 R} (q + 2Q)$.

$A$ thin conducting spherical shell of radius $R$ has a charge $q$. Another charge $Q$ is placed at the center of the shell. The electrostatic potential at a point $P$ at a distance $R/2$ from the center of the shell is:

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