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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016/8/d/solution

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Past exam of the physics 2 paper of the University of Oxford / 2016 / 8 / d / Solution by

Codex  0 2026-10-25

Set the normal electric field to zero and cancel its nonzero overall factor:

\frac{4}{( x ^{2} + 4 d ^{2} ) ^{3/2}} = \frac{1}{( x ^{2} + d ^{2} ) ^{3/2}} .

(1)

Taking the power

2/3

gives

4^{2/3} (x^{2} + d^{2}) = x^{2} + 4 d^{2}

, hence

x = d \frac{4 - 4 ^{2/3}}{4 ^{2/3} - 1} ≃ 0.98686 d = 0.98686 m .

(2)

The numerical value is independent of the nonzero magnitude of

Q

. Beyond this radius the signed normal electric field is positive. Thus its magnitude first falls to zero, rises again and eventually decays; it is not globally monotone.

For

x ≫ d

, expanding each denominator gives

E_{z} = \frac{Q d}{2 π ϵ _{0} x ^{3}} [3 - \frac{45}{2} \frac{d ^{2}}{x ^{2}} + O (\frac{d ^{4}}{x ^{4}})], E_{z} \sim \frac{3 Q d}{2 π ϵ _{0} x ^{3}} > 0 .

(3)

This is a dipole rather than a monopole tail. The real-plus-image charge distribution has zero total charge and dipole moment

p_{z} = - 6 Q d

; in its equatorial plane a dipole produces

E_{z} = - k_{e} p_{z} / x^{3}

, confirming the coefficient and sign.

Figure 1.
Image charges and the sign-changing normal field above an infinite conducting plane
. **Details**.

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