A thin spherical conducting shell of radius R | vedclass

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(D) The total electrostatic potential at point $P$ is the sum of the potential due to the point charge $Q$ at the centre and the potential due to the

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The total electrostatic potential at point $P$ is the sum of the potential due to the point charge $Q$ at the centre and the potential due to the spherical shell of charge $q$.$1$. The potential due to the point charge $Q$ at a distance $r = \frac{R}{2}$ is given by $V_Q = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{R/2} = \frac{2Q}{4\pi \varepsilon_0 R}$.$2$. For a spherical conducting shell of radius $R$ with charge $q$,the potential at any point inside the shell (including the centre) is constant and equal to the potential at its surface. Thus,the potential at distance $r = \frac{R}{2}$ due to the shell is $V_q = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{R}$.$3$. The total potential at point $P$ is $V = V_Q + V_q = \frac{2Q}{4\pi \varepsilon_0 R} + \frac{q}{4\pi \varepsilon_0 R}$.Therefore,the correct option is $(d)$.

$A$ 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 $\frac{R}{2}$ from the centre of the shell is

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