A thin spherical conducting shell of radius R | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The potential at point $P$ inside the shell is the sum of the potential due to the charge $Q$ at the centre and the potential due to the charge $q$ on the shell.$1$. The potential due to charge $Q$ at a distance $r = R/2$ is given by $V_Q = \frac{1}{4\pi \epsilon_0} \frac{Q}{R/2} = \frac{2Q}{4\pi \epsilon_0 R}$.$2$. The potential due to the spherical shell of charge $q$ at any point inside it (including the centre) is constant and equal to the potential at its surface,which is $V_q = \frac{1}{4\pi \epsilon_0} \frac{q}{R}$.$3$. The total potential at point $P$ is $V_P = V_Q + V_q = \frac{2Q}{4\pi \epsilon_0 R} + \frac{q}{4\pi \epsilon_0 R}$.

$A$ 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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