Figure shows a solid hemisphere with a charge of 5nC distributed uniformly throughout its volume. The hemisphere lies on a plane and point P is located on the plane, along a radial line from the centre of curvature at distance 15 cm. The electric potential(volts) at point P due to the hemisphere, is: $(\frac{1}{4 π ε_{0}} = 9 \times 10^{9} (S . I u n i t s))$

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answer is 300.

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Detailed Solution

By argument of symmetry, it will be half of the potential produced by the full sphere

$∵$ Charge on hemisphere $= Q$ , so charge on sphere $= 2 Q$

$\Rightarrow V = \frac{1}{2} . \frac{K (2 Q)}{R} = \frac{K Q}{R}$

$V = \frac{K Q}{R} = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{15 \times 10^{- 2}} = 300 V$

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