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Deduce the Expression for the Potential Energy of a System of Two Charges - Physics

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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.

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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`

Concept: Potential Energy in an External Field - Potential Energy of a System of Two Charges in an External Field
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