Applying Gauss' Law: Planar Symmetry — Question 1

Question 1

An isolated insulating sheet splits its flux equally between two sides

An infinite nonconducting sheet in $x=0$ carries uniform density $\sigma>0$. There are no other sources. Choose a symmetric pillbox with each end face area A.

1. Derive the field magnitude and direction on each side.

2. Compare this result with the field immediately outside a conductor surface having the same local density. Explain the factor of two by counting the nonzero pillbox faces.

Question 1 - Solution

An isolated insulating sheet splits its flux equally between two sides

(a) Method and calculation

$$\Phi=EA+EA=2EA,\quad Q_{\mathrm{enc}}=\sigma A.$$$$\mathbf E(x)=\frac\sigma{2\epsilon_0}\operatorname{sgn}(x)\hat{\mathbf x}\quad(x\ne0).$$

(b) Interpretation and further calculation

$$\text{Conductor:}\quad E_{\mathrm{metal}}=0,\quad E_{\mathrm{vacuum}}A=\sigma A/\epsilon_0.$$$$E_{\mathrm{vacuum}}=\sigma/\epsilon_0=2E_{\mathrm{isolated\ sheet}}.$$The single-sheet result relies on reflection symmetry; the conductor boundary instead has zero field on its metal side.

Physical check

The factor of two is not a contradiction: the field on the second side differs in the two physical systems.