Problem 14: Show that the plates of a parallel-plate capacitor attract each other with a force given by

$$ F=\frac{q^2}{2\epsilon A} $$

Prove this by calculating the work necessary to increase the plate separation form x to dx, the charge q remains constant

Note: A Common Mistake

It is tempting to use Guass’ law and the equation $F=qE$ to calculate electrical force. If using the approach, electric field between two parallel plates is equal to $E=q/(2A\epsilon_0)$. Then, electrical force $F$ is equal to:

$$ F=\frac{q^2}{\epsilon_0 A} $$

which is off by a factor $1/2$. Why $F=qE$ gives a different answer?

• The expression $F=qE$ typically describes a single point/test charge $q$ in an electric field $E$. When dealing with the force between the plates of a capacitor, we need to consider the interaction between the distributed charges on the plate. While $E={q}/{A\epsilon_0}$ is the total electric field between plates, the qE approach does not account for the distribution of forces over the entire area of plates. In other words, $F=qE$ (or Coulomb’s law) applies for point charges and not continuous distribution of charges over an area, explaining why the answer is wrong.

Problem 15:

Solution:

From problem 14, we know that the electrical force between two parallel plates ie equal to $F=q^2/(2\epsilon_0 A)$. Then, force per area is equal to:

$$ \frac{F}{A}=\frac{q^2}{2\epsilon_0 A^2}=\frac{\sigma^2}{2\epsilon_0} $$

By Guass’ law, we know that the electric field between two parallel plates is $E=\sigma/\epsilon_0$, giving us $\sigma=\epsilon{E}$. Then, the equation becomes:

$$ \boxed{\frac{F}{A}=\frac{\sigma^2}{2\epsilon}=\frac{1}{2}\epsilon_0 E^2} $$