A thin spherical conducting shell of radius R | vedclass

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Question

(D) The potential at point $P$ inside the shell is the sum of the potential due to the charge $Q$ at the center and the potential due to the charge $q

Vedclass · Accepted Answer

$\frac{2Q}{4 \pi \varepsilon_0 R} + \frac{q}{4 \pi \varepsilon_0 R}$

Vedclass · Answer

(D) The potential at point $P$ inside the shell is the sum of the potential due to the charge $Q$ at the center and the potential due to the charge $q$ on the shell.For a point inside a conducting shell,the potential due to the shell's charge $q$ is constant and equal to the potential at its surface,which is $V_{shell} = \frac{kq}{R}$.The potential due to the point charge $Q$ at distance $r = R/2$ is $V_{point} = \frac{kQ}{R/2} = \frac{2kQ}{R}$.Therefore,the total potential at point $P$ is $V_P = V_{point} + V_{shell} = \frac{2kQ}{R} + \frac{kq}{R}$.Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $V_P = \frac{2Q}{4 \pi \varepsilon_0 R} + \frac{q}{4 \pi \varepsilon_0 R}$.

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

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