Question

A sphere of radius 10 mm carries a charge of 8nc is attached at the end of an insulating dtring

Answer 1

Answer:

The average current is $1.27\times10^{-5}\ A$.

Explanation:

Suppose that, A sphere of radius 10 mm carries a charge of 8 nc is attached at the end of an insulating string and is whirled in a circle. The rotation frequency is 100 rad/s.

The average current does this rotating charge represent will be

We need to calculate the surface area of the sphere

$A=4\pi r^2$

$A=4\times\pi\times(10\times10^{-3})^2$

$A=0.001256\ m^2$

We need to calculate the charge density

Using formula of charge density

$\delta=\dfrac{q}{A}$

Put the value into the formula

$\delta=\dfrac{8\times10^{-9}}{0.001256}$

$\delta=6.36\times10^{-6}\ C/m^2$

We need to calculate the current

Using formula of current

$I=\int_{0}^{\pi}{\omega r\sin\theta\ \delta}$

Now to get the total current we have to integrate the above equation from 0 to π.

$I_{tot}=\omega r\delta\int_{0}^{\pi}{\sin\theta}$

$I_{tot}=2\omega r\delta$

Put the value into the formula

$I_{tot}=2\times100\times10\times10^{-3}\times6.36\times10^{-6}$

$I_{tot}=1.27\times10^{-5}\ A$

Hence, The average current is $1.27\times10^{-5}\ A$.