A Charged Isolated Conductor — Question 5

Question 5

A small pillbox turns a local surface field into a patch charge

Use $\epsilon_0=8.854\times10^{-12}\,\mathrm{C^2/(N\,m^2)}$. Near a smooth conductor surface in electrostatic equilibrium, a small vacuum-side probe measures $\mathbf E=2.00\times10^5\hat{\mathbf n}$ N/C, where $\hat{\mathbf n}$ points from metal into vacuum. Treat the field as uniform over a patch of area $3.00\times10^{-6}$ m$^2$.

1. Derive the local relation between field and surface charge density, then estimate the patch charge.

2. Explain why a nonzero tangential static field would contradict equilibrium, and determine the direction of force on a positive weak probe just outside.

Question 5 - Solution

A small pillbox turns a local surface field into a patch charge

(a) Method and calculation

$$\Phi=E_n\Delta A,\quad Q_{\mathrm{enc}}=\sigma\Delta A\Rightarrow\sigma=\epsilon_0E_n.$$$$\sigma=1.7708\times10^{-6}\,\mathrm{C/m^2},\quad Q_{\mathrm{patch}}=\sigma\Delta A=5.3124\,\mathrm{pC}.$$

(b) Interpretation and further calculation

$$\mathbf F_\parallel=q_{\mathrm{mobile}}\mathbf E_\parallel,\quad\text{equilibrium}\Rightarrow\mathbf E_\parallel=\mathbf0.$$$$q_t>0:\quad\mathbf F_t=q_tE_n\hat{\mathbf n}.$$A tangential field would drive surface charges to redistribute; the positive probe force here points away from the metal.

Physical check

The patch-charge estimate assumes the measured normal field does not vary appreciably across the stated small area.