Answer:
The charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] 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 :
[tex]V=\dfrac{kq}{r}[/tex]
q is the charge on the sphere
[tex]q=\dfrac{Vr}{k}[/tex]
[tex]q=\dfrac{200\times 0.15}{9\times 10^9}[/tex]
[tex]q=3.34\times 10^{-9}\ C[/tex]
Charge per unit area is called charge density on the surface. it is given by :
[tex]\sigma=\dfrac{q}{A}[/tex]
[tex]\sigma=\dfrac{q}{4\pi r^2}[/tex]
[tex]\sigma=\dfrac{3.34\times 10^{-9}}{4\pi (0.15)^2}[/tex]
[tex]\sigma=1.18\times 10^{-8}\ C/m^2[/tex]
So, the charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively. Hence, this is the required solution.
