Consider an arbitrary charged conductor with a local surface charge density $\sigma$. The electric f

Question

Consider an arbitrary charged conductor with a local surface charge density $\sigma$. The electric force experienced per unit area of the conductor's surface (also known as electrostatic pressure) is given by:

Professor PC Thomas Classes · Answer

The electric field just outside a conductor is $E_{out} = \frac{\sigma}{\epsilon_0}$, and inside it is $E_{in} = 0$. A small surface element $dA$ carrying charge $dq = \sigma dA$ does not exert a force on itself. The force on it is due to the electric field created by the *rest* of the charges on the conductor. This field is the average of the field just outside and just inside: $E_{rest} = \frac{1}{2}(E_{out} + E_{in}) = \frac{\sigma}{2\epsilon_0}$. The force $dF$ on the element is $dq \cdot E_{rest} = (\sigma dA) \frac{\sigma}{2\epsilon_0} = \frac{\sigma^2}{2\epsilon_0} dA$. Therefore, the force per unit area is $P = \frac{dF}{dA} = \frac{\sigma^2}{2\epsilon_0}$.