Question

A charged belt, 61 cm wide, travels at 43 m/s between a source of charge and a sphere. The belt carries charge into the sphere at a rate corresponding to 100 µA. Compute the surface charge density on the belt.

Answer 1

Answer:

Surface charge density will be $3.812\times 10^{-6}C/m^2$

Explanation:

We have given speed of the belt v = 43 m/sec

Width of the belt w = 61 cm

We know that charge is equal to Q = It, here Is current and t is time

And time is equal to $i=\frac{L}{v}$, here L is distance and v is speed

Putting the value of t in charge equation

$Q=i\times \frac{L}{v}=\frac{iL}{v}$

Surface charge density is equal to $\sigma =\frac{Q}{A}=\frac{iL}{vA}$

We know that width is equal to $w=\frac{A}{L}$

So $\sigma =\frac{i}{vw}=\frac{100\times 10^{-6}}{43\times 0.61}=3.812\times 10^{-6}C/m^2$