Question

A conducting shell of radius R is cut in half. The two hemispherical pieces are electrically separated from each other but left close together, so that the distance separating the two halves can be neglected. The upper half is maintained at a constant potential V-Vo, and the lower half is grounded (V0).

a) Determine the boundary condition at r R as a function of θ and expand it in a series of Legendre polynomials, retaining the terms up to l = 3 only.
b) Calculate the electrostatic potential V at all points outside of the surface of the conductors. Neglect terms falling faster than 1/r

Answer 1

The boundary condition at r = R can be determined by applying the fact that the electric field must be continuous across the boundary of the two hemispheres. The electrostatic potential V at all points outside the surface of the conductors can be calculated using the method of images.

Step-by-step explanation:

The boundary condition at r = R can be determined by applying the fact that the electric field must be continuous across the boundary of the two hemispheres. Therefore, the electric field on the upper half is E = -(V - V0)/(Rsin(θ)), where θ is the angle measured from the z-axis. Expanding this expression in a series of Legendre polynomials and retaining terms up to l = 3, we get: E = -(V - V0)/R * (1 - (3/2)sin^2(θ) + (15/8)sin^4(θ))

The electrostatic potential V at all points outside the surface of the conductors can be calculated using the method of images. The potential at a point P with distance r from the origin and angle θ can be found by summing the potentials due to the positive and negative charges in the system. The potential due to the positive charge is V1 = kQ/r, and the potential due to the negative charge is V2 = -kQ/r, where k is the Coulomb's constant. Therefore, the total potential at point P is V = V1 + V2 = kQ((1/r) - (1/r'))