Question

what are the charge and the charge density on the surface of a conducting sphere of radius 0.15 m whose potantial is 200v?

Answer 1

Answer:

The charge and the charge density on the surface of a conducting sphere is $3.34\times 10^{-9}\ C$ and $1.18\times 10^{-8}\ C/m^2$ respectively.

Explanation:

It is given that,

Radius of the conducting sphere, r = 0.15 m

Potential, V = 200 V

Potential on the surface of sphere is given by :

$V=\dfrac{kq}{r}$

q is the charge on the sphere

$q=\dfrac{Vr}{k}$

$q=\dfrac{200\times 0.15}{9\times 10^9}$

$q=3.34\times 10^{-9}\ C$

Charge per unit area is called charge density on the surface. it is given by :

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

$\sigma=\dfrac{q}{4\pi r^2}$

$\sigma=\dfrac{3.34\times 10^{-9}}{4\pi (0.15)^2}$

$\sigma=1.18\times 10^{-8}\ C/m^2$

So, the charge and the charge density on the surface of a conducting sphere is $3.34\times 10^{-9}\ C$ and $1.18\times 10^{-8}\ C/m^2$ respectively. Hence, this is the required solution.