Question

A hollow spherical conductor of radius 0.2m has a charge of 4nC. Find the potential due to this charged sphere on its surface, inside the surface and at a distance of 0.25m outside from its surface.​

Answer 1

31 is the answer but not shore

Answer 2

Given:

Radius of hollow spherical conductor = 0.2m

Charge of hollow spherical conductor = 4nC

To find:

Potential due to charge on the surface, inside the surface and at the distance 0.25cm outside the surface.

Solution:

Formula used to find the potential due to charge will be shown below,

                       $V = \frac{k \times q}{r}$

here, V = potential

        k = $\frac{1}{4\pi \epsilon_o }$ = 9 × 10⁹

        q = charge

        r = radius of spherical conductor

according to the question,

   q = 4nC

    R = radius of spherical conductor = 0.2m

     r = distance outside the surface = 0.25m

Now, calculate the potential on the surface when r = R, we get,

           $V = \frac{1}{4\pi \epsilon_o } \times \frac{q}{R}$

          $V = \frac{9 \times 10^9 \times 4}{0.2}$

         V = 180 × 10⁹

Now, calculate the potential inside the surface when R > r, we get,

         $V = \frac{1}{4\pi \epsilon_o } \times \frac{q}{R}$

         $V = \frac{9 \times 10^9 \times 4}{0.2}$

        V = 180 × 10⁹

Now, calculate the potential outside the surface when R < r, we get,

           $V = \frac{1}{4\pi \epsilon_o } \times \frac{q}{R}$

           $V = \frac{9 \times 10^9 \times 4}{0.25}$

          V = 144 × 10⁹

So, the potential on the surface is 180 × 10⁹, inside the surface is 180 × 10⁹ and at a distance of 0.25m outside the surface is 144 × 10⁹.