 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022/9/a/solution

Set the grounded plane at

z = 0

and the real point charge at

(0, 0, h)

- q

(0, 0, - h)

makes the electrostatic potential vanish throughout the plane. The uniqueness theorem for the electrostatic potential then identifies the physical solution in

z > 0

Φ (a, z) = \frac{q}{4 π ϵ _{0}} [\frac{1}{a ^{2} + ( z - h ) ^{2}} - \frac{1}{a ^{2} + ( z + h ) ^{2}}] .

The normal electric field is

E_{z} = - \partial_{z} Φ

E_{z} (a, z) = \frac{q}{4 π ϵ _{0}} [\frac{z - h}{[ a ^{2} + ( z - h ) ^{2} ] ^{3/2}} - \frac{z + h}{[ a ^{2} + ( z + h ) ^{2} ] ^{3/2}}],

and therefore

E_{z} (a, 0^{+}) = - \frac{q}{2 π ϵ _{0}} \frac{h}{( a ^{2} + h ^{2} ) ^{3/2}} .

For

q > 0

the electric field points into the electrical conductor, consistently with negative induced surface charge density.

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