Question

Use the on-axis potential of a charged disk from Chapter 25 to find the on-axis electric field of a charged disk

Short answer

The electric field of charged disk is$E\left(z\right)=-\frac{Q}{2\mathrm{\pi }{\in }_0}{R}^2\frac{z}{\sqrt{R^2+z^2}}-1$

Step-by-step solution

Given information

We need to find the on-axis electric field of a charged disk

Simplify

The potential on the axis of a disk with charge $Q$and radius $R$at distance $z$from it is given as

$V\left(z\right)=\frac{Q}{2\mathrm{\pi }{\in }_0{\mathrm{R}}^2}\sqrt{R^2+z^2-z}$.

The formula for electric field is

$E=-\frac{dV}{dr}$,

Where $r$is the direction in which we're investigating the electric field. We have to take the negative of the derivative of the first expression with respect to $z$. It is given as

$E=-\frac{dV}{dz}=-\frac{Q}{2\mathrm{\pi }{\in }_0{\mathrm{R}}^2}\frac{d}{dz}\sqrt{R^2+z^2-z}$

If we perform the calculation correctly we get

$E\left(z\right)=-\frac{Q}{2\mathrm{\pi }{\in }_0{\mathrm{R}}^2}\frac{z}{\sqrt{R^2+z^2}}-1$