Question

A charged belt, 50 cmwide, travels at 30m/sbetween a source of charge and a sphere. The belt carries charge into the sphere at a rate corresponding to$\mathbf{100}\mathbf{\mu A}$. Compute the surface charge density on the belt.

Short answer

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

Step-by-step solution

The given data

1. Width of the belt, $\mathrm{w}=50\mathrm{cm}\mathrm{or}0.50\mathrm{m}$
2. Speed of belt, v = 30 m/s
3. Current rate,$\mathrm{i}=100\mathrm{\mu A}\mathrm{or}100\times {10}^{-6}\mathrm{A}$

Understanding the concept of the surface charge density

Surface charge density is the amount of charge stored on the surface per unit area. It is measured in Coulomb per meter squared.

To find the surface charge density on the belt, we have to use the expression for the area of the belt in the formula of surface charge density.

Formulae:

The charge density of the material, $\sigma =\frac{q}{A}$ ...(i)

The area of the cross-section of the material to the flow rate, A = wvt ...(ii)

The current passing through the area in terms of charge, i = q/t ...(iii)

Calculation of the surface charge density

Substituting the value for area and charge from equations (ii) and (iii) in the equation of charge density of equation (i), we get the require value of the surface charge density of the material as follows:

$$\begin{gathered} \mathrm{\sigma }=\frac{\mathrm{it}}{\mathrm{wvt}} \\ =\frac{100\times {10}^{-6}\mathrm{A}}{0.50\mathrm{m}\times 30\mathrm{m}/\mathrm{s}} \\ =6.66\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \\ \approx 6.7\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2\mathrm{C}/\mathrm{m} \end{gathered}$$

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