Calculating the Potential from the Field — Question 6

Question 6

Mixed coordinate dependence can still give a path-independent voltage

A planar field is $\mathbf E=\beta y\hat{\mathbf x}+\beta x\hat{\mathbf y}$ with constant $\beta$. Set V(0,0)=0.

1. Integrate along the coordinate axes to obtain V(a,b).

2. Repeat along the straight path $(x,y)=(at,bt)$, $0\le t\le1$, and find a potential valid everywhere.

Question 6 - Solution

Mixed coordinate dependence can still give a path-independent voltage

(a) Method and calculation

$$\int_{(0,0)}^{(a,0)}\mathbf E\cdot d\mathbf l=0,\quad\int_{(a,0)}^{(a,b)}\mathbf E\cdot d\mathbf l=\beta ab.$$$$V(a,b)=-\beta ab.$$

(b) Interpretation and further calculation

$$\Delta V=-\int_0^1[\beta(bt)a+\beta(at)b]dt=-\beta ab.$$$$V(x,y)=-\beta xy,\quad-\nabla V=\beta y\hat{\mathbf x}+\beta x\hat{\mathbf y}.$$

Physical check

The equality of the cross derivatives of the field components supports the conservative result in this simply connected domain.