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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2018/1/solution

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Past exam of the physics 2 paper of the University of Oxford / 2018 / 1 / Solution by

Codex  0 2026-10-25

The electric field of a signed point electric charge in vacuum follows Coulomb's law:

E (r) = \frac{Q}{4 π ϵ _{0}} \frac{r}{∣ r ∣ ^{3}} .

For the conducting plane, the method of images replaces each charge

Q_{j}

h_{j}

by an image charge

- Q_{j}

- h_{j}

. The real and image electric fields have cancelling tangential components and equal normal components at the surface. Thus the electric field at a conducting plane from axial point charges, just above the point directly below them, is

E_{z} (0^{+}) = - \frac{2}{4 π ϵ _{0}} \sum_{j} \frac{Q _{j}}{h _{j}^{2}} = \frac{60}{4 π ϵ _{0}} (\frac{1}{500 0 ^{2}} - \frac{1}{800 0 ^{2}}) .

Numerically,

E (0^{+}) ≃ 1.31 \times 1 0^{4} V m^{- 1} \overset{z}{^}, upwards .

The surface value means the air-side limit. Inside the ideal conductor the electrostatic electric field is zero; its discontinuity across the surface is supported by induced surface charge. Omitting the image charges would give half the external answer.

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