# Question #a0b33

Nov 26, 2017

$F = {q}^{2} / \left(2 {\epsilon}_{0}\right)$

#### Explanation:

The total force on a group of charges with total charge Q is given by

$\textcolor{b l u e}{F = Q E}$

Here we will treat the charge of each plate like the total charge Q.

Before we can find the force on each plate, we must find the electric field of each of the plates. A simple method for doing so makes use of Gauss's law.

• Model each of the plates like an infinite plane carrying a charge. We will look at the top, positively charged plate.

• We can then draw an appropriate Guassian surface, such as a Guassian "pill box." This should extend equal distances above and below the plane.

• Apply Gauss's Law to this surface:

$\textcolor{b l u e}{\oint \text{ "vecE*dveca=1/(epsilon_0)Q_"enclosed}}$

In this case, ${Q}_{\text{enc}} = + q$ for the positively charged plate. By symmetry, $\vec{E}$ points away from the plane—upward for points above, downward for points below.

• For the negatively charged plate, the field points toward it.

So, the top and bottom surfaces of the top plate yield:

$\textcolor{b l u e}{\int \text{ } \vec{E} \cdot \mathrm{dv} e c a = 2 A \left\mid \vec{E} \right\mid}$

whereas the sides contribute nothing (no flux through these regions as the field is normal to the surface). Thus:

$2 A \left\mid \vec{E} \right\mid = \frac{1}{{\epsilon}_{0}} q$

$\implies \textcolor{b l u e}{\vec{E} = \frac{q}{2 {\epsilon}_{0}} \hat{n}}$

where $\hat{n}$ is a unit vector pointing away from the surface. This depends on how the plates are oriented (your choice of/given coordinate system).

Then the field for the bottom plate is the same.

• Field due to bottom plate: $\textcolor{b l u e}{\vec{E} = - \frac{q}{2 {\epsilon}_{0}} \hat{n}}$

Between a parallel plate capacitor, the electric fields of each plate point in the same direction (toward the negatively charged plate), so the field adds (note that the normal vector is different for each plate, hence the opposite sings). However, outside of the plates (both above and below), the fields cancel.

Each plate does not exert a force on itself, so we only consider the field produced by one plate.

The force of attraction between the two plates is therefore:

$F = Q E$

$\implies \textcolor{b l u e}{F = {q}^{2} / \left(2 {\epsilon}_{0}\right)}$