 Solution

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

1/ r^{2}

throughout the vacuum gap, so its maximum occurs just outside the inner sphere:

E_{m a x} = \frac{∣ V _{1} ∣ b}{a ( b - a )} = \frac{1 0 ^{3} \times 0.4}{0.1 ( 0.4 - 0.1 )} .

Thus

E_{m a x} = 1.333 \times 1 0^{4} V m^{- 1}, r = a^{+} .

Inside the metal the electric field is zero. The one-sided gap value is the relevant value for dielectric breakdown; the ideal zero-thickness conducting boundary itself should not be assigned a single field value across its surface-charge discontinuity.

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