Question

What is the differential relationship between the magnitude of the electric field \[|E|\] and the potential gradient along the normal displacement \[dl\]?

Options

• \[|E| = \left[\frac{dl}{dV}\right]\]

• \[|E| = \left[\frac{dV}{dl}\right]^2\]

• \[|E| = -\left[\frac{dV}{dl}\right]\]

• \[|E| = \int V \, dl\]

Answer 1

Equating the work done \[|E|\delta l\] to the potential difference \[V_A - V_B = -\delta V\] yields \[|E| = -\left[\frac{\delta V}{\delta l}\right]\]. In differential form, this gives the potential gradient relation \[|E| = -\left[\frac{dV}{dl}\right]\].