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Deduce the expression for the potential energy of a system of two point charges q_1 and q_2 brought from infinity to the points vecr_1 and vecr_2 respectively in the presence of external electric field vecE. |
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Answer» Solution :To deduce the potential energy of a system of two point charges `q_(1)` and `q_(2)` being brought from infinity to the points `vecr_(1)` and `vecr_(2)` respectively in an external electric field `vecE_(1)`, first we calculate the WORK DONE in bringing the CHARGE `q_1` from infinity of `vecr_1` . Work done in this step is `W_(1) = q_(1) V (vecr_(1))` , where `V (vecr_(1))` is the electrostatic potential at `r_1` . Now we consider the work done in bringing `q_2` to `vecr_2` . In this step work `W_(2)` is done against the electric field and work `W_(12)` is done against the field DUE to charge `q_(1)` . Obviously `W_(2) = q_(2) V (vecr_(2))` where `V (vecr_(2))` is the electrostatic potential at `vecr_(2)` and `W_(12) = (q_(1) q_(2))/(4 pi in_(0) . r_(12))` , where `r_(12) = |vecr_(1) - vecr_(2)|` = distance between `q_(1)` and `q_(2)` `therefore` Total work done in bringing `q_(2)` to `vecr_(2) = W_(2) + W_(12) = q_(2) V (vecr_(2)) + (q_(1) q_(2))/(4pi in_(0) . r_(12))` `therefore` Potential energy of the system U = The total work done in assembling the configuration `U = W_(1) + W_(2) + W_(12) = q_(1) V (r_(1)) + q_(2) V (r_(2)) + (q_(1) q_(2))/(4 pi in_(0) . r_(12))` |
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