An ideal air-filled parallel-plate capacitor with horizontal plates has a plate separation of $5.0 \mathrm {~cm}$. If the potential difference between the plates is $2000 \mathrm {~V}$, with the top plate at the higher potential, what are the magnitude and direction of the electric field between the plates?

Question

Quizplus · Accepted Answer

The electric field $E$ between the plates of a parallel-plate capacitor is given by $E = \frac{V}{d}$, where $V$ is the potential difference and $d$ is the separation between the plates. Here, $E = \frac{2000 \mathrm{~V}}{0.05 \mathrm{~m}} = 40000 \mathrm{~N/C}$. The field direction is from the higher potential to the lower potential, which is downward.

An Ideal Air-Filled Parallel-Plate Capacitor with Horizontal Plates Has a Plate