Electric potential on the surface of a solid sphere of charge is V, Radius of the sphere is 1m Match the following two columns,

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Solution

Given , $V = \frac{k q}{R} = k q$ as $R = 1$
A) $V_{R / 2} = \frac{k q}{R} - \frac{k q}{R^{3}} \int_{R}^{R / 2} r d r = \frac{k q}{R} + \frac{3 k q}{8 R} = \frac{11 k q}{8 R} = \frac{11}{8} V$
B) $V_{2 R} = \frac{k q}{2 R} = \frac{V}{2}$
C) using Gauss's law, $E . 4 π r^{2} = \frac{ρ . (4 / 3 π r^{3})}{ϵ_{0}} = \frac{q / R^{3}}{ϵ_{0}}$ where $q = (4 / 3) π R^{3}$
or $E = \frac{1}{4 π ϵ_{0}} \frac{q r}{R^{3}} = k \frac{q r}{R^{3}}$
thus $E_{R / 2} = k \frac{q}{2 R^{2}} = \frac{V}{2}$ (as $R = 1, V = k q$ )
D) $E_{2 R} = \frac{k q}{(2 R)^{2}} = \frac{k q}{4 R^{2}} = \frac{V}{4}$ (as $R = 1, V = k q$ )

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Electric potential on the surface of a solid sphere of charge is V, Radius of the sphere is 1m Match the following two columns,

Given , V=kqR=kq as R=1 A) VR/2=kqR−kqR3∫R/2Rrdr=kqR+3kq8R=11kq8R=118VB) V2R=kq2R=V2 C) using Gauss's law, E.4πr2=ρ.(4/3πr3)ϵ0=q/R3ϵ0 where q=(4/3)πR3or E=14πϵ0qrR3=kqrR3thus ER/2=kq2R2=V2 (as R=1,V=kq)D) E2R=kq(2R)2=kq4R2=V4 (as R=1,V=kq)