A thin spherical insulating shell of radius R | vedclass

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Question

(A) Let the initial charge on the sphere be $Q$. Thus,$V_0 = \frac{kQ}{R}$.When a small hole of area $\alpha(4\pi R^2)$ is made,the charge removed is

Vedclass · Accepted Answer

The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2} R$ from the center towards the hole is $\frac{1-\alpha}{1-2\alpha}$.

Vedclass · Answer

(A) Let the initial charge on the sphere be $Q$. Thus,$V_0 = \frac{kQ}{R}$.When a small hole of area $\alpha(4\pi R^2)$ is made,the charge removed is $q = \alpha Q$.$(1)$ Potential at the center $(V_c)$ and at a point $P$ at distance $R/2$ from the center towards the hole $(V_p)$:The potential at any point is the sum of the potential due to the complete sphere and the potential due to the removed charge element (treated as a negative point charge at the surface).$V_c = \frac{kQ}{R} - \frac{kq}{R} = \frac{kQ}{R}(1-\alpha) = V_0(1-\alpha)$.$V_p = \frac{kQ}{R} - \frac{kq}{R/2} = \frac{kQ}{R} - \frac{2kq}{R} = \frac{kQ}{R}(1-2\alpha) = V_0(1-2\alpha)$.Therefore,the ratio $\frac{V_c}{V_p} = \frac{1-\alpha}{1-2\alpha}$. Thus,option $(A)$ is correct.$(2)$ Electric field at the center $(E_c)$:Initially,$E_c = 0$. After removing charge $q$,$E_c = \frac{kq}{R^2} = \frac{k(\alpha Q)}{R^2} = \alpha \frac{V_0}{R}$. The field increases.$(3)$ Electric field at a point $P'$ at distance $2R$ from the center:Initially,$E_{P'} = \frac{kQ}{(2R)^2} = \frac{kQ}{4R^2}$.After removing charge $q$ from the surface,the field due to the hole at $P'$ is $\frac{kq}{R^2}$ (since $P'$ is at distance $R$ from the hole).$E_{P', 	ext{final}} = \frac{kQ}{4R^2} - \frac{kq}{R^2} = \frac{kQ}{4R^2} - \frac{k(\alpha Q)}{R^2} = \frac{kQ}{4R^2} - \frac{4k\alpha Q}{4R^2} = \frac{kQ}{4R^2}(1-4\alpha)$.Change in field $\Delta E = \frac{kq}{R^2} = \frac{k(\alpha Q)}{R^2} = \frac{\alpha V_0}{R}$.

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