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A uniformly charged solid hemisphere of radius b and charge density rho has a hemispherical cavity of radius a = b/2 scooped out from its centre. If the electric potential at the centre of the cavity is n*rho*b² / (16*epsilon₀), find n.

1. 1
2. 2
3. 3
4. 4

Correct answer: 3

Solution

For a uniformly charged solid sphere of radius R with charge density rho, the potential at centre = 3*rho*R² / (6*epsilon₀) *... Let's compute carefully. For a full solid sphere: V(centre) = (3/2) * (rho * R²)/(3*epsilon₀) = rho*R²/(2*epsilon₀). For a hemisphere: by symmetry considerations, potential at the flat face centre of a hemisphere = half that of full sphere? Actually, potential is a scalar and has contributions from all charges. For a solid hemisphere of radius R, charge density rho, potential at the centre of the flat face: integrate over the hemisphere. This is complex. Using superposition principle: hemisphere = full sphere - other hemisphere. By symmetry, potential at centre of each hemisphere = potential at centre of full sphere/2... No, potential is scalar: V(centre of sphere) = V due to upper hemisphere + V due to lower hemisphere. By symmetry both contribute equally, so V(centre of sphere due to one hemisphere) = V(centre of sphere)/2. V(centre of full solid sphere) = rho/(epsilon₀) * R²/2... Standard result: V(centre of solid sphere) = (3/2) * (1/(4*pi*epsilon₀)) * (4/3*pi*R³*rho)/R = (3*rho*R²)/(6*epsilon₀) = rho*R²/(2*epsilon₀). Contribution from each hemisphere = rho*R²/(4*epsilon₀). Now for the problem: original hemisphere radius b, cavity hemisphere radius a=b/2. V at cavity centre = V(large hemisphere, radius b, at its flat centre) - V(small hemisphere, radius a=b/2, at its flat centre). V_large = rho*b²/(4*epsilon₀). V_small = rho*a²/(4*epsilon₀) = rho*(b/2)²/(4*epsilon₀) = rho*b²/(16*epsilon₀). V = rho*b²/(4*epsilon₀) - rho*b²/(16*epsilon₀) = rho*b²*(4-1)/(16*epsilon₀) = 3*rho*b²/(16*epsilon₀). So n = 3.

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