A Charged Isolated Conductor — Question 7

Question 7

Surface pressure requires removing a patch from its own field

A small patch on a conductor in electrostatic equilibrium has surface density $\sigma$. Locally the field just outside is $\sigma\hat{\mathbf n}/\epsilon_0$. Model the patch self-field as $\sigma\hat{\mathbf n}/(2\epsilon_0)$ just outside; the remaining field is due to the other charges.

1. Derive the outward electric force per unit area on the patch and show that its direction is independent of the sign of $\sigma$.

2. For an isolated spherical conductor of radius R and charge Q, integrate the z component of this pressure over its upper hemisphere to find the separating force on that hemisphere.

Question 7 - Solution

Surface pressure requires removing a patch from its own field

(a) Method and calculation

$$\mathbf E_{\mathrm{other}}=\frac{\sigma}{\epsilon_0}\hat{\mathbf n}-\frac{\sigma}{2\epsilon_0}\hat{\mathbf n}=\frac{\sigma}{2\epsilon_0}\hat{\mathbf n}.$$$$\frac{d\mathbf F}{dA}=\sigma\mathbf E_{\mathrm{other}}=\frac{\sigma^2}{2\epsilon_0}\hat{\mathbf n},\quad P_e=\frac{\epsilon_0E_s^2}{2}.$$

(b) Interpretation and further calculation

$$F_z=P_e\int_{\mathrm{hemisphere}}\cos\theta\,dA=P_e\pi R^2.$$$$\sigma=\frac Q{4\pi R^2}\Rightarrow F_z=\frac{Q^2}{32\pi\epsilon_0R^2}.$$The vector force integral uses projected area, not the full hemisphere area.

Physical check

A charge patch cannot exert a net force on itself; using the full exterior field would give twice the correct local pressure.