# Electrostatic potential Part – 4

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, VS â€“ V = 0 Â V = VS].

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 = -Edrcos00 = – dr or, V = = –

Or, V – = – [r2 â€“ R2]

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 VA = . 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 VB = . As the sphere A is inside the sphere B, now the potential of sphere A is = = Â = VA.

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 q1 and q2.

As q1 charge remain on outer surface of shell A then no charge is on the inner surface of it.

Let q3 and q4 are the charge of inner and outer surface of shell B respectively. Using Gaussâ€™s law through the material of B we get, q3 = – q1.

Let q5 and q6 are the charge of inner and outer surface of shell C. Using Gaussâ€™s law through the material of C we get, q5 = – q4. So, q2 = q5 + q6.

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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