A thin spherical conducting shell of radius R | vedclass

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(D) The potential at any point inside a spherical conducting shell is constant and equal to the potential at its surface.Given,charge on the shell is

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The potential at any point inside a spherical conducting shell is constant and equal to the potential at its surface.Given,charge on the shell is $q$ and its radius is $R$.The potential due to the shell at any point inside (including the centre and at distance $R/2$) is $V_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{R}$.The potential due to the point charge $Q$ placed at the centre at a distance $r = R/2$ is $V_2 = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R/2} = \frac{2Q}{4\pi\varepsilon_0 R}$.The total electrostatic potential at point $P$ is $V = V_1 + V_2$.Therefore,$V = \frac{q}{4\pi\varepsilon_0 R} + \frac{2Q}{4\pi\varepsilon_0 R} = \frac{1}{4\pi\varepsilon_0 R} (q + 2Q)$.

$A$ thin spherical conducting shell of radius $R$ has 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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