Question

An insulating belt moves at speed 30 m/sand has a width of 50 cm. It carries charge into an experimental device at a rate corresponding to 100 mA. What is the surface charge density on the belt?

Short answer

The surface charge density on the belt is$6.7\times {10}^{-6}\frac{\mathrm{C}}{{\mathrm{m}}^2}$ .

Step-by-step solution

The given data

1. Velocity of the belt, v = 30 m/s
2. Width of the belt, b = 50 cm or 0.5m
3. Current through the belt,$\mathrm{I}=100\mathrm{\mu A}\mathrm{or}100\times {10}^{-6}\mathrm{A}$

Understanding the concept of the surface charge density

By using the formula for current density and adjusting it for surface charge density, we will find the solution.

Formulae:

The total charge present in the volume,$q=nALe$ (i)

The current density of a material having a drift of electrons, $J=\frac{I}{A}=ne{v}_d$ (ii)

Calculation of the surface charge density

The surface charge density of the present electrons can be calculated using equation (i) as follows:

$\frac{q}{A}=nLe...............................................................................\left(1\right)$

Area of the conveyor belt is given by, Area = Length (L) ×Width (b)

Thus, from the current density equation (ii), we can get the value of the surface charge using the given data and equation (1) as follows:

$$\begin{gathered} \frac{\mathrm{I}}{\mathrm{L}\times \mathrm{b}}=\mathrm{n}\mathrm{e}{\mathrm{v}}_{\mathrm{d}}(\because \mathrm{A}=\mathrm{L}\times \mathrm{b}) \\ \mathrm{L}\mathrm{n}\mathrm{e}=\frac{\mathrm{I}}{{\mathrm{bv}}_{\mathrm{d}}} \\ =\frac{100\times {10}^{-6}}{0.5\times 30} \\ =6.66\times {10}^{-6} \\ =6.7\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the charge density is$6.7\mathrm{x}{10}^{-6}\mathrm{C}/{\mathrm{m}}^2$$6.7\mathrm{x}{10}^{-6}\mathrm{C}/{\mathrm{m}}^2$.