 Solution

ID: past-exam-of-the-electromagnetism-course-of-the-university-of-oxford/2023/1/a/iii/solution

Q = 2 π a^{2} \frac{q ( a - r _{0}^{2} / a )}{4 π} \int_{- 1}^{1} \frac{d u}{( a ^{2} + r _{0}^{2} - 2 a r _{0} u ) ^{3/2}} .

Since

r_{0} > a

, the elementary integral is

\frac{1}{a r _{0}} [\frac{1}{a ^{2} + r _{0}^{2} - 2 a r _{0} u}]_{- 1}^{1} = \frac{1}{a r _{0}} (\frac{1}{r _{0} - a} - \frac{1}{r _{0} + a}) = \frac{2}{r _{0} ( r _{0}^{2} - a ^{2} )} .

Substitution yields the total induced electric charge

Q = - \frac{q a}{r _{0}} = q^{⋆} .

The equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.

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