Deduce the expression for the potential energy of a system of two charges q_{1} and q_{2} located `vec(r_1)` and `vec(r_2)`, respectively, in an external electric field.

#### Solution

Let q₁ and q₂ be the two charges located at r₁ and r₂, respectively, in an external electric field. The work done in bringing the chare q₁ from infinity to r₁ is W₁ = q₁V (r₁), where V(r₁) is the potential.

Similarly, the work done in bringing the chare q₁ from infinity to r₂ can be calculated. Here, the work is done not only against the external field E but also against the field due to q₁.

Hence, work done on q₂ against the external field is W₂ = q₂V (r₂).

Work done on q against the field due to q_{1}, W_{12} = `(q_1q_2)/(4piin_0r_12`where r₁₂ is the distance between q₁ and q₂.

By the principle of superposition for fields, work done on q₂ against two fields will add with work done in bringing q₂ to r₂, which is given as

`W_2+W_12=q_2V(r_2)+(q_1q_2)/(4piin_0r_12)`

Thus, the potential energy of the system U = total work done in assembling the configuration U = W₁ + W₂ + W + ₁₂

`U=q_1V(r_1)+q_2V(r_2)+(q_1q_2)/(4piin_0r_12`