Question

The electric field just above the surface of the charged conducting drum of a photocopying machine has a magnitude Eof$\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$ . What is the surface charge density on the drum?

Short answer

The surface charge density of the drum is $2.0\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2$.

Step-by-step solution

The given data

The magnitude of the electric field,$\left|\mathrm{E}\right|=2.3\times {10}^5\mathrm{N}/\mathrm{C}$

Understanding the concept of Gauss law-planar symmetry

Using the concept of the electric field of a conducting sheet, we can calculate the required surface charge density of the drum.

Formula:

The electric field of a conducting sheet, $\mathrm{E}=\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0}$ (1)

Calculation of the surface charge density

Using the given data in equation (1), we can get the value of the surface charge density of the drum such that,

$$\begin{gathered} \mathrm{\sigma }={\mathrm{E\epsilon }}_0 \\ =\left(2.3\times {10}^5\mathrm{N}/\mathrm{C}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right) \\ =2.0\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the required value of the density is $2.0\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2$.