Question

A 5 cm radius isolated conducting sphere is charged so its potential is 100 V, relative to the potential far away. The charge density on its surface is:

Answer 1

The electric potential (V) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation:

$V = \frac{kQ}{r}$

Here,

k = Coulomb's constant

Q = Charge

r = Distance

If we rearrange the equation to find the distance we have,

$Q=\frac{Vr}{k}$

$Q = \frac{(100)(5*10^{-2})}{9*10^9}$

$Q = 5.55*10^{-10} C$

Now the value of the charge density is equivalent to the charge on the surface area of the sphere this is

$\gamma = \frac{Q}{A_s}$

$\gamma = \frac{Q}{4\pi r^2}$

$\gamma = \frac{ 5.55*10^{-10}}{4\pi (5*10^{-2})^2}$

$\gamma = 1.76*10^{-8} C/m^2$