Question

The figure below shows a solid hemisphere with a charge of $5nC$ distributed uniformly throughout its volume. The hemisphere lies on a plane and point is located on the plane, along a radial line from the centre of curvature at a distance of 15 cm. The electric potential at point due to the hemisphere is V

Answer 1

300 V

Answer 2

The problem asks for electric potential due to a uniformly charged solid hemisphere at an external point. As the radius of the hemisphere is not provided, the most reasonable approach is to approximate the hemisphere as a point charge located at its center of curvature (center of its base). The potential due to a point charge is $V = kQ/r$. Substituting the given charge $Q = 5 \times 10^{-9} \, C$ and distance $r = 0.15 \, m$, and Coulomb's constant $k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2$, the potential is calculated as $V = \frac{9 \times 10^9 \times 5 \times 10^{-9}}{0.15} = 300 \, V$.