Electric potential: The electric potential at a point in the electric field due to a charge is the amount of work done to bring a unit positive charge (test charge) from infinity to that point against the electrostatics force of the electric field of source charge.

Units of electrostatic potential:

C.G.S. unit: statvolt – If one erg of work is required to move charge of one statcoulomb from infinity to the given point, against the electrostatic ﬁeld, then the potential at the given point is 1 statvolt. So, 1 statvolt = .

SI unit: volt – If one joule of work is required to move charge of one coulomb from infinity to the given point, against the electrostatic ﬁeld, then the potential at the given point is 1volt. So, 1 volt = .

Relation between volt and statvolt: 1 volt = = = statvolt.

In C.G.S. unit electromagnetic system, the unit of electrostatic potential is emu of potential or abvolt. So, 1 volt = = 10^{8} abvolt.

Potential at a point due to a point charge:

Let us consider a point charge q is placed at point O. To calculate electric potential at a point P at a distance r from O in the electric field of q, we consider a test charge q_{0} is moved from infinity to point P.

The force acting on q_{0} when it is at moved from infinity is = ———-(i)

To move q_{0} at constant speed an external force = is applied in opposite direction.

The work done by F to move q_{0} from P^{/} to P^{//} (where P^{/}P^{//} = dr^{/}) is dw = – dr^{/} ——–(ii)

[-ve is due to the work done by external force F to move q_{0} towards q.]

From equations (i) and (ii) we get, dw = –

The total work done by external force F to move q_{0} from infinity to point P is W = dw = – = – kqq_{0}

= – kqq_{0}[ – ] =

The potential at a point P is V(q_{0} = 1) = = = .

_{1}and q

_{2}placed at a certain distance r is U = .

Potential energy of a system of three charges:

Let us consider three point charges q_{1}, q_{2}, and q_{3} are placed at infinity. To calculate potential energy at point A, B and C due to charges q_{1}, q_{2}, and q_{3} respectively, we have to bring the charges from infinity to those points.

As initially, there is no charge at point A, the potential at point A, V_{A} = 0. So the work done to bring charge q_{1} from infinity to point A is W_{A} = V_{A}.q_{1} = 0.

The potential at point B due to q_{1} is V_{B} = .

So the work done to bring charge q_{2} from infinity to point B is W_{B} = V_{B}.q_{2} = .

The potential at point C due to q_{1} and q_{2} is V_{C} = + .

So the work done to bring charge q_{3} from infinity to point C is W_{C} = V_{C}.q_{3} = +

The potential energy due to the system of three charges is W = W_{A }+ W_{B} + W_{C}

= [ + + ].

Relation between electric potential and electric field:

Let us consider two points P and Q are very close to each other, separated by a distance dr in electric field along the direction .

The force acting on the test charge q_{0} placed at Q is = q_{0} ——(1)

The work done by external force to move q_{0} from Q to P is dw = – .^{ }——–(2)

[-ve is due to the work done by external force to move q_{0} against .]

From equations (1) and (2) we get, dw = – q_{0}.

Or, = – .

So, dV = – . (the work done dw to move q_{0 }in potential difference dV between points P and Q is dw = q_{0}.dV.)

The electric field is equal to the negative of the potential gradient.

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