D.2.9—Parallel-plate field

Use the uniform-field model

Between two large opposite parallel plates, away from the edges, field strength equals potential difference $V$ divided by perpendicular plate separation $d$. The field points from the positive plate to the negative plate; edge regions are not uniform.

Worked example — required potential difference

For $E=1.0\times10^6\,\mathrm{V\,m^{-1}}$ and $d=0.50\,\mathrm{cm}=5.0\times10^{-3}\,\mathrm{m}$, $V=Ed=(1.0\times10^6)(5.0\times10^{-3})=5.0\times10^3\,\mathrm{V}$. The result applies to the uniform central region.

Read the direction

Electric field lines point from the positive plate to the negative plate, so a positive charge accelerates in that direction and a negative charge accelerates oppositely. The field strength can be expressed in $\mathrm{N\,C^{-1}}$ or equivalently $\mathrm{V\,m^{-1}}$.

Check the boundary

The formula assumes a uniform region and neglects edge effects. Use the perpendicular plate separation in metres; do not use the diagonal distance or the plate length.

Common trap

Do not reverse the field direction because the test charge is negative. Field direction is defined by a positive test charge; the force on a negative charge is opposite.