Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 3 / d

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- Solution d

Solution

Differentiate the Green-function representation, valid away from the charged sheet:

E_{y} = \frac{1}{2 π ϵ _{0}} \int_{0}^{L} \frac{y d s}{( x - s ) ^{2} + y ^{2}} = \frac{a r c t a n ( x / y ) - a r c t a n (( x - L ) / y )}{2 π ϵ _{0}} .

For

0 < x < L

, the two inverse tangent limits at

y ↓ 0

are

π /2

and

- π /2

; approaching from below reverses both signs. Thus

E_{y} (x, 0^{+}) = \frac{1}{2 ϵ _{0}}, E_{y} (x, 0^{-}) = - \frac{1}{2 ϵ _{0}}, ∣Δ E_{y} ∣ = \frac{1}{ϵ _{0}} .

This agrees with the normal electric field jump obtained by applying Gauss's law to a thin pillbox enclosing unit surface charge density.

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