Question

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

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

Answer 1

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

Let $\sigma$ is the surface charge density. Length entering per unit time into the sphere $0.30m/s$.

The charge is $dq=Idt$

The surface is

$dA=Ldx$

charge density is

$\sigma =\frac{dq}{dA}$

$\sigma =\frac{Idt}{Ldx}$

$\sigma =\frac{I}{L\frac{dx}{dt}}$

$\sigma =\frac{I}{Lv}$

$\Rightarrow \sigma \left(0.15\right)={10}^{-4}\Rightarrow \sigma =\frac{100}{15}\times {10}^{-4}=6.7\times {10}^{-4}C/m^{-2}$