Question

The belt of an electrostatic generator is $50cm$ wide and travels $30cm/s$. The belt carries charge into the sphere at a rate corresponding to ${10}^{-4}A$. What is the surface density of charge on the belt?

• A. $6.7\times {10}^{-5}C{m}^{-2}$
• B. $6.7\times {10}^{-7}C{m}^{-2}$
• C. $6.7\times {10}^{-4}C{m}^{-2}$
• D. $6.7\times {10}^{-8}C{m}^{-2}$

Answer 1

The correct option is C $6.7\times {10}^{-4}C{m}^{-2}$

Let the surface charge density of the belt be $\sigma$.
Velocity of the belt, $V=30cm/s=0.3m/s$
Width of the belt, $w=50cm=0.5m$
Since surface charge density, $\sigma =\frac{Q}{A}$
$Q=\sigma \times A$
Dividing both sides by time $t$
$\begin{array}{}\frac{Q}{t}=\frac{\sigma \times A}{t}& \text{(1)}\end{array}$
Since $\frac{Q}{t}=I$ and $A=l\times w$
where $l$ is the length of the belt
$\Rightarrow \frac{A}{t}=\frac{l\times w}{t}=V\times w$
Therefore equation (1) becomes,
$I=\sigma \times V\times w$
${10}^{-4}=\sigma \times 0.3\times 0.5$
$\Rightarrow \sigma =6.67\times {10}^{-4}C{m}^{-2}\approx 6.7\times {10}^{-4}C{m}^{-2}$