Equipotential Surfaces — Question 4

Question 4

Two branches of the same equipotential may meet at a critical point

A local potential is $V(x,y)=A(x^2-y^2)$ with $A>0$, independent of z. A sketch shows two V=0 lines crossing at the origin.

1. Find both zero-potential branches and the field at their intersection.

2. Find the field at (a,a) and explain why this crossing does not allow two different voltage values at one point.

Question 4 - Solution

Two branches of the same equipotential may meet at a critical point

(a) Method and calculation

$$V=0\Rightarrow y=x\ \text{or}\ y=-x.$$$$\mathbf E=-2Ax\hat{\mathbf x}+2Ay\hat{\mathbf y},\quad\mathbf E(0,0)=\mathbf0.$$

(b) Interpretation and further calculation

$$\mathbf E(a,a)=2Aa(-\hat{\mathbf x}+\hat{\mathbf y}),\quad\mathbf E\cdot(\hat{\mathbf x}+\hat{\mathbf y})=0.$$$$V(0,0)=0\quad\text{on both branches}.$$Different-valued level surfaces cannot intersect; equal-valued branches can meet where the gradient vanishes.

Physical check

At a critical point the zero field has no unique direction, so the usual nonzero normal-vector construction does not apply.