A solid spherical non-conductor of radius 5 cm has the potential of 50 V at its outer surface. The potential at its center is

100 v

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75 v

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Zero

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50 v

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Solution

The correct option is B 75 v
$V (r) = \frac{k Q}{2 R} (R^{2})$
So,
$V (R) = \frac{k Q}{2 R} \times 2 V (R) = \frac{k Q}{R}$
Given, $V (R) = 50 R = 5 c m = 0.05 m 50 = k Q / 0.05 k Q = 2.5 \to (1)$
Now potential at centre of surface
$V (r = 0) = k Q / 2 R (3.0 / R^{2}) = 3 k Q / 2 R$
From $(1)$
$V (r = 0) = 3 \times 2.5 / 2 \times 0.05 = 75 v$

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A solid spherical non-conductor of radius 5 cm has the potential of 50 V at its outer surface. The potential at its center is

The correct option is B 75 vV(r)=kQ2R(R2)So,V(R)=kQ2R×2V(R)=kQRGiven, V(R)=50R=5cm=0.05m50=kQ/0.05kQ=2.5→(1)Now potential at centre of surfaceV(r=0)=kQ/2R(3.0/R2)=3kQ/2RFrom (1)V(r=0)=3×2.5/2×0.05=75v