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Electric potential on the surface of a uniformly charged solid sphere is V. Radius of the sphere is R(=1m). Match the following two columns. (‘r’ is the distance from the centre of the sphere)
Column – I Column – II
A Electric potential at $r = \frac{R}{2}$ P $\frac{V}{4}$
B Electric potential at $r = 2 R$ Q $\frac{V}{2}$
C Electric field at $r = \frac{3 R}{4}$ R $\frac{3 V}{4}$
D Electric field at $r = 2 R$ S $\frac{11 V}{8}$

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A – S; B – Q; C – R; D – P

A – P; B – R; C – Q; D – S

A – R; B – Q; C – P; D – S

A – S; B – P; C – Q; D - R

answer is C.

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Detailed Solution

$V = \frac{K Q}{2 R^{3}} [3 R^{2} - r^{2}]$

At $r = \frac{R}{2}, V = \frac{K Q}{2 R^{3}} (3 R^{2} - \frac{R^{2}}{4}) = \frac{11 K Q}{8 Q} = \frac{11 V}{8}$

$V = \frac{K Q}{2 R} = \frac{V}{2}$

$E = \frac{K Q}{R^{3}} \frac{R}{2} = \frac{K Q}{2 R^{2}} = \frac{V}{2}$

$E = \frac{K Q}{4 R^{2}} = \frac{V}{4}$

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