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

Surface charge density will be
`3.812* 10^(-6)C/m^2`

Step-by-step 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=(L)/(v)`, here L is distance and v is speed

Putting the value of t in charge equation

`Q=i* (L)/(v)=(iL)/(v)`

Surface charge density is equal to
`\sigma =(Q)/(A)=(iL)/(vA)`

We know that width is equal to
`w=(A)/(L)`

So
`\sigma =(i)/(vw)=(100* 10^(-6))/(43* 0.61)=3.812* 10^(-6)C/m^2`