A uniformly charged solid sphere of radius R | vedclass

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(B) The potential on the surface of a uniformly charged solid sphere is $V_0 = \frac{Kq}{R}$.For $r < R$, the potential is $V_i = \frac{Kq}{2R^3}(3R^2

Vedclass · Accepted Answer

$R_1 = 0$ and $R_2 < (R_4 - R_3)$

Vedclass · Answer

(B) The potential on the surface of a uniformly charged solid sphere is $V_0 = \frac{Kq}{R}$.For $r At the center $(r = 0)$, $V_{center} = \frac{3Kq}{2R} = \frac{3}{2}V_0$. Thus, $R_1 = 0$ for potential $\frac{3V_0}{2}$.For potential $\frac{5V_0}{4}$ $(r For potential $\frac{3V_0}{4}$ $(r > R)$: $\frac{3}{4} \frac{Kq}{R} = \frac{Kq}{R_3} \implies R_3 = \frac{4}{3}R \approx 1.333R$.For potential $\frac{V_0}{4}$ $(r > R)$: $\frac{1}{4} \frac{Kq}{R} = \frac{Kq}{R_4} \implies R_4 = 4R$.Now, $R_2 = 0.707R$ and $(R_4 - R_3) = 4R - 1.333R = 2.667R$.Since $0.707R < 2.667R$, we have $R_2 < (R_4 - R_3)$.

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