Solution

Inserting an independently biased conducting sphere makes two vacuum gaps, each with its own constant radial electric flux. Let

A_{1}

and

A_{2}

1/ r^{2}

. The two potential drops are

V_{1} - V_{2} = A_{1} (\frac{1}{a} - \frac{1}{c}), V_{2} = A_{2} (\frac{1}{c} - \frac{1}{b}) .

Therefore

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

The middle conductor can carry net electric charge, so

A_{1}

and

A_{2}

need not be equal. Its charge is

4 π ϵ_{0} (A_{2} - A_{1})

, consistent with the jump in the normal electric field. For

0 < V_{2} < V_{1}

both fields point outwards; outside that interval one of the gap fields reverses direction.

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