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Question
Deduce the expression for the potential energy of a system of two charges q1 and q2 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 q1, W12 = `(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`
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