Question

The electric field just above the surface of the charged drum of a photocopying machine has a magnitude E of 2.3 × 105 N/C. What is the surface charge density on the drum, assuming that the drum is a conductor?
_____________ C/m^2

Answer 1

Concepts and reason

The concepts used to solve this problem are electric field and charge density.

Use the permittivity of free space and electric field to find the surface charge density.

Fundamentals

The expression for the electric field in terms of charge density is given by the expression as follows:

$E = \frac{\sigma }{{{\varepsilon _o}}}$

Here, $E$ is the electric field, $\sigma$ is the charge density and ${\varepsilon _0}$ is the permittivity of free space.

The expression for the electric field in terms of charge density is given by the expression as follows:

$E = \frac{\sigma }{{{\varepsilon _o}}}$

Rearrange the above expression to obtain the surface charge density.

$\sigma = {\varepsilon _o}E$

The expression for the surface charge density is as follows:

$\sigma = {\varepsilon _o}E$

Substitute $2.3 \times {10^5}\,{\rm{N/C}}$ for $E$ and $8.854 \times {10^{ - 12}}\,{\mathop{\rm F}\nolimits} /m$ for ${\varepsilon _0}$ .

$\begin{array}{c}\\\sigma = \left( {8.854 \times {{10}^{ - 12}}\,{\mathop{\rm F}\nolimits} /m} \right)\left( {2.3 \times {{10}^5}\,{\rm{N/C}}} \right)\\\\ = 2.036 \times {10^{ - 6}}\,C/{m^2}\\\end{array}$

Ans:

The surface charge density is ${\bf{2}}{\bf{.036 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\,{\bf{C/}}{{\bf{m}}^{\bf{2}}}$ .