Question

A very small earthed conducting sphere is at a distance $a$ from a point charge ${\dot{q}}_{1}$ and at a distance $b$ from a point charge $q_{2} (a < b)$ . At a certain instant, the sphere starts expanding so that its radius grows according to the law $R = ν t$ . Determine the time dependence $I (t)$ of the current in the earthing conductor, assuming that the point charges and the centre of the sphere are at rest, and in due time the initial point charges get into the expanding sphere without touching it (through small holes).

Accepted Answer

We shall write the condition of the equality to zero of the potential of the sphere, and hence of any point inside it (in particular, it's center), by the moment of time, $t$ . We shall single out three-time intervals:

(1) $t < \frac{a}{v}$ ,

(2) $\frac{a}{v} ⩽ t < \frac{b}{v}$

(3) $I (t)$

Denoting the charge of the sphere by, $q (t)$ , we obtain the following expression for an instant $t$ from the first time interval:

$\frac{q_{1}}{a} + \frac{q_{2}}{b} + \frac{q (t)}{v t} = 0$ ,

hence,

$q (t) = - v (\frac{q_{1}}{a} + \frac{q_{2}}{b}) t_{1}$ ,

$I_{1} (t) = - v (\frac{q_{1}}{a} + \frac{q_{2}}{b})$ .

For an instant $t$ from the second time interval, we find that the fields inside and outside the sphere are independent, and hence

$\frac{q (t) + q_{1}}{v t} = - \frac{q_{2}}{b}, I_{2} (t) = - v \frac{q_{2}}{b}$ .

Finally, as soon as the sphere absorbs the two-point charges $q_{1}$ and $q_{2}$ , the current will stop flowing through the "earthing" conductor, and we can write $I_{3} (t) = 0$ .

$I (t) = \{\begin{array}{lc} - v (\frac{q_{1}}{a} + \frac{q_{2}}{b}), & t < \frac{a}{v} \\ - v \frac{q_{2}}{b}, & \frac{a}{v} ⩽ t < \frac{b}{v} \\ 0, & t ⩾ \frac{b}{v} \end{array}$

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