1. Electric potential due to uniformly charged spherical shell: Let us consider a spherical shell of radius R is charged by q. Calculate electric field at a point P, at a distance r from the centre of the shell.

(i) If point P is outside the shell then potential at P is V = .

(ii) If point P is on the surface of shell then potential at P is V = .

(iii) If point P is inside the shell or at the centre of the shell then potential at P is V = .

[Inside the spherical shell E = 0, therefore, – Â = Â = 0 or, V_{S} â€“ V = 0 Â V = V_{S}].

2. Electric potential due to uniformly charged solid sphere: Let us consider a solid sphere of radius R is charged by q. Calculate electric field at a point P, at a distance r from the centre of the sphere.

(i) If point P is outside the sphere then potential at P is V = .

(ii) If point P is on the surface of sphere then potential at P is V = .

(iii) If point P is inside the sphere then using Gaussâ€™s law the electric field E = .

Now dV = -Edrcos0^{0} = – dr or, V = = –

Or, V – = – [r^{2} â€“ R^{2}]

Or, V = [1 + Â – ] = [ – ].

3. Concept on potential due to shearing of charge: A solid sphere (A) of radius a is charged by q and its potential is V_{A} = . Now the sphere is connected with a hollow sphere (B) of radius b using a conducting wire. Now the charge will reside on the outer surface of B [using Gaussâ€™s law].

The potential of B is V_{B} = . As the sphere A is inside the sphere B, now the potential of sphere A is = = Â = V_{A}.

4. Charge appears on different surfaces: Let us consider three spherical shells A, B and C of radii a, b and c respectively. Shell B is grounded and shell A and C are charged by q_{1} and q_{2}.

As q_{1 }charge remain on outer surface of shell A then no charge is on the inner surface of it.

Let q_{3} and q_{4} are the charge of inner and outer surface of shell B respectively. Using Gaussâ€™s law through the material of B we get, q_{3} = – q_{1}.

Let q_{5} and q_{6} are the charge of inner and outer surface of shell C. Using Gaussâ€™s law through the material of C we get, q_{5} = – q_{4}. So, q_{2} = q_{5} + q_{6}.

As the potential of shell B is 0 then k[Â + Â + ] = 0.

Using these conditions charge of different surfaces can be calculated.

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