Electrostatic potential Part – 3

Concepts on electrostatic potential:

1. Electrostatic potential at a point on the axis of a uniformly charged ring:

Let us consider a ring of radius R is uniformly charged by q. To calculate electrostatic potential at a point P on the axis of ring at a distance y from the centre O, we consider an elementary charge dq. The electrostatic potential at point P due to dq is dV = .

The electrostatic potential at a point P due to the entire ring is V =

=Â  Â = = Â [where k = ].

2. Electrostatic potential at a point on the axis of a uniformly charged disc:

Let us consider a disc of radius R is uniformly charged by q. To calculate electrostatic potential at a point P on the axis of disc at a distance y from the centre O, we consider an elementary ring of radius r thickness dr with charge dq.

So, dq = Â = . The electrostatic potential at a point P due to dq is dV = Â = .

The electrostatic potential at a point P due to the entire disc is

V = Â = =

[let r2 + y2 = z2 differentiating both side with z we get, rdr = zdz therefore, Â = Â = Â = Â â€“ y]

At the centre of the disc (y = 0) V = = [where k = ].

When y >> R then Â = y[1 + ]1/2 = y + Â [using binomial theorem]

So, V = [y + – y] = Â = .

3. Potential at a point on the rim of a uniformly charged disc: A disc of radius a is uniformly charged by q. P is a point on the edge of the disc. We consider an elementary area at an angle with diameter of the disc. The charge of the elementary area rddr on the surface of the disc is dq = rddr.

The electrostatic potential due to dq at point P is dV = Â = .

Potential at point P due to the disc is V = Â =

From diagram rm = PA = 2acos

Â V = Â = .

4. Relation between electric potential and electric field: If uniform electric field (= EcosiÌ‚ + EsinjÌ‚) is directed at an angle Î¸ with +ve X axis then find the potential difference between point A and B.

Here = -aiÌ‚ + bjÌ‚. So potential difference between point A and B is VA â€“ VB = – Â = –.(dxiÌ‚ + dyjÌ‚) = – dx – dy

Or, VA â€“ VB = E[acos + bsin].

5. Conservation of energy: A charge particle q is shot towards another charge particle Q which is fixed, with a speed v. It approaches Q upto a closest distance r and then returns. The kinetic energy of charge q = Potential energy of charge q and Q at closest approach. So, mv2 = .

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