Calculating the Field from the Potential — Question 9

Question 9

A localized voltage profile gives two extrema of electric-field magnitude

Along x, $V(x)=V_0e^{-x^2/a^2}$ with $V_0,a>0$.

1. Find E(x), its zeros and its direction on either side of the origin.

2. Find the positions and values of the largest field magnitude.

Question 9 - Solution

A localized voltage profile gives two extrema of electric-field magnitude

(a) Method and calculation

$$E_x=-V\prime=\frac{2V_0x}{a^2}e^{-x^2/a^2}.$$$$E_x(0)=0,\quad\operatorname{sgn}(E_x)=\operatorname{sgn}(x).$$

(b) Interpretation and further calculation

$$\frac{dE_x}{dx}=\frac{2V_0}{a^2}e^{-x^2/a^2}\left(1-\frac{2x^2}{a^2}\right)=0\Rightarrow x=\pm a/\sqrt2.$$$$|E|_{\max}=\frac{\sqrt2V_0}{a}e^{-1/2}.$$

Physical check

The largest voltage occurs where the field vanishes; largest field magnitude occurs where the voltage slope is steepest.