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Derive an expression for potential energy of a system of two charges in the absence of the external electric field. |
Answer» Solution : Imagine the charges `q_1` and `q_2`are initially at infinity. First, the charge `q_1`is brought from infinity to the point A and no work is done for this. This is because there is no electric FIELD to oppose this charge. Consider another point B at a distance from A. Electric potential at B due to r is given by, `V_1 = (1)/(4pi epsi_0 )(q_1)/(r )` Next the charge `q_2` is brought from infinity to the point B. When the charge 92 is MOVED, the electric field due to `q_1` opposes it. Hence work has to be done. Work done in bringing the charge `q-2`from infinity to the point B. `W = V_1 xx q_2` [since `V = (W)/(q_0) rArr W = qV`] `W = (1)/(4pi epsi_0) (q_1)/(r ) xx q_2` This work done is stored in the charges as electric potential energy and given by `PE=U = (1)/(4pi epsi_0) (q_1 q_2)/(r )` |
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