 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2016/9/a/solution

E = A \hat{r} / r^{2}

in the gap, where

A

is constant. Integrating the electric field between the two equipotentials gives

V_{1} = V (a) - V (b) = \int_{a}^{b} \frac{A}{r ^{2}} d r = A (\frac{1}{a} - \frac{1}{b}) .

Consequently

E (r) = \frac{V _{1} ab}{( b - a ) r ^{2}} \hat{r}, a < r < b .

The inner charge is

Q = 4 π ϵ_{0} A

4 π ϵ_{0} ab / (b - a)

. The radial direction is outward for positive

V_{1}

and inward for negative

V_{1}

; the formula represents the signed field, not just its magnitude.

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