Equipotential Surfaces — Question 10

Question 10

Normal spacing changes along one parabolic equipotential family

A two-dimensional potential is $V=A(y-bx^2)$, with $A>0$ in V/m and $b>0$ in inverse meters. Consider nearby levels separated by small $|\delta V|$.

1. Find their shapes and the local electric field.

2. Derive their normal spacing at x and its ratio to the spacing at x=0; evaluate the ratio at $x=1/(2b)$.

Question 10 - Solution

Normal spacing changes along one parabolic equipotential family

(a) Method and calculation

$$V=V_c\Rightarrow y=bx^2+V_c/A.$$$$\mathbf E=2Abx\hat{\mathbf x}-A\hat{\mathbf y},\quad|\mathbf E|=A\sqrt{1+4b^2x^2}.$$

(b) Interpretation and further calculation

$$\delta n\simeq\frac{|\delta V|}{A\sqrt{1+4b^2x^2}},\quad\frac{\delta n(x)}{\delta n(0)}=\frac1{\sqrt{1+4b^2x^2}}.$$$$x=1/(2b):\quad\delta n(x)/\delta n(0)=1/\sqrt2.$$

Physical check

The vertical separation is constant, but the relevant distance for field strength is along the local normal.