Question

Coulomb’s law says that electric field falls off like $\mathbf{1}\mathbf{/}{\mathit{z}}^{\mathbf{2}}$. How can $\mathit{E}$for a uniformly charged disk depend on $\left[1-z/R\right]$, or be independent of distance?

Short answer

Electric field can be independent of distance but is inversely proportional to the area of the disk which has dimension $L^2$.

Step-by-step solution

Given data

Electric field due to a uniformly charged disk is independent of distance.

Field due to a uniformly charged disk

The electric field on the axis of a uniformly charged disk of charge $\mathit{Q}$ and $\mathit{A}\mathbf{}$area extremely close to the disk is

$\mathit{E}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{2}\mathbf{A}{\mathbf{\epsilon }}_{\mathbf{0}}}$ ......(i)

Here, localid="1668493266480" ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity of free space

Determination of the dependence of electric field on length dimension

The field given in equation (i) might be explicitly independent of distance but is inversely proportional to area which has dimension

$\left[A\right]=L^2$

Thus the dimension remains correct.