Potential of a Charged Isolated Conductor — Question 4

Question 4

Charging a sphere requires integrating its increasing potential

An initially neutral isolated conducting sphere of radius R is charged quasistatically to Q>0 by bringing infinitesimal charge from infinity. Ignore dissipative losses.

1. Derive its charging work and explain the factor one half relative to Q times the final voltage.

2. Now expand the isolated charged sphere quasistatically to radius 2R with Q fixed. Find the change in electrostatic energy and the electrical contribution to external work.

Question 4 - Solution

Charging a sphere requires integrating its increasing potential

(a) Method and calculation

$$V(q\prime)=kq\prime/R,\quad W=\int_0^Q V(q\prime)dq\prime=\frac{kQ^2}{2R}.$$$$W=\tfrac12QV_f,\quad V_f=kQ/R;\quad\overline V=V_f/2.$$

(b) Interpretation and further calculation

$$U_i=kQ^2/(2R),\quad U_f=kQ^2/(4R).$$$$\Delta U=W_{\mathrm{ext,electrical}}=-kQ^2/(4R).$$Mechanical material energies of the expanding sphere are excluded.

Physical check

The energy falls upon expansion at fixed charge; using Q times the final voltage would count every pair interaction twice.