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(D) The potential at the surface of the conductor is $V_s = \frac{q}{4\pi\varepsilon_0 r}$.The potential at a distance $3r$ from the center is $V_p =

Vedclass · Accepted Answer

$\frac{V}{6r}$

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(D) The potential at the surface of the conductor is $V_s = \frac{q}{4\pi\varepsilon_0 r}$.The potential at a distance $3r$ from the center is $V_p = \frac{q}{4\pi\varepsilon_0 (3r)}$.The potential difference $V$ is given by $V = V_s - V_p = \frac{q}{4\pi\varepsilon_0 r} - \frac{q}{12\pi\varepsilon_0 r} = \frac{q}{4\pi\varepsilon_0 r} \left(1 - \frac{1}{3}\right) = \frac{2q}{12\pi\varepsilon_0 r} = \frac{q}{6\pi\varepsilon_0 r}$.Thus,$\frac{q}{4\pi\varepsilon_0} = \frac{3}{2} Vr$.The electric intensity $E$ at a distance $3r$ is $E = \frac{q}{4\pi\varepsilon_0 (3r)^2} = \frac{q}{4\pi\varepsilon_0 (9r^2)}$.Substituting the value of $\frac{q}{4\pi\varepsilon_0}$,we get $E = \frac{3Vr}{2(9r^2)} = \frac{V}{6r}$.

$A$ charged spherical conductor has radius '$r$'. The potential difference between its surface and a point at a distance '$3r$' from the center is '$V$'. The electric intensity at a distance '$3r$' from the center of the conductor is:

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