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(B) The potential at the surface of the sphere is $V_s = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.The potential at a distance $5R$ from the centre is $V_

Vedclass · Accepted Answer

$\frac{V}{20R}$

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(B) The potential at the surface of the sphere is $V_s = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$.The potential at a distance $5R$ from the centre is $V_p = \frac{1}{4\pi\epsilon_0} \frac{q}{5R}$.The potential difference $V$ is given by $V = V_s - V_p = \frac{q}{4\pi\epsilon_0} (\frac{1}{R} - \frac{1}{5R}) = \frac{q}{4\pi\epsilon_0} (\frac{4}{5R}) = \frac{q}{5\pi\epsilon_0 R}$.From this,we can express the charge $q$ as $q = \frac{5\pi\epsilon_0 RV}{1}$.The electric field intensity $E$ at a distance $r = 5R$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{q}{(5R)^2} = \frac{q}{100\pi\epsilon_0 R^2}$.Substituting the value of $q$ in terms of $V$: $E = \frac{1}{100\pi\epsilon_0 R^2} \cdot (5\pi\epsilon_0 RV) = \frac{5}{100} \frac{V}{R} = \frac{V}{20R}$.

$A$ hollow charged metal sphere has radius $R$. If the potential difference between its surface and a point at a distance $5R$ from the centre is $V$,then the magnitude of the electric field intensity at a distance $5R$ from the centre of the sphere is:

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