+q charges are placed at a, 3a, 5a… (up to infinity) and –q charges are placed at 2a, 4a, 6a….. (up to infinity). What is the total potential energy at origin (0, 0)?(a) Zero(b) \(\frac {q}{a}\)(c) \(\frac {(q \, log2)}{a}\)(d) \(\frac {qe^2}{a}\)The question was asked in exam.I'm obligated to ask this question of Potential Energy of a System of Charges in chapter Electrostatic Potential and Capacitance of Physics – Class 12

+q charges are placed at a, 3a, 5a… (up to infinity) and –q charge

1.	+q charges are placed at a, 3a, 5a… (up to infinity) and –q charges are placed at 2a, 4a, 6a….. (up to infinity). What is the total potential energy at origin (0, 0)?(a) Zero(b) \(\frac {q}{a}\)(c) \(\frac {(q \, log2)}{a}\)(d) \(\frac {qe^2}{a}\)The question was asked in exam.I'm obligated to ask this question of Potential Energy of a System of Charges in chapter Electrostatic Potential and Capacitance of Physics – Class 12
Answer» Correct answer is (c) \(\FRAC {(q \, LOG2)}{a}\) Best explanation: The total potential energy at the origin DUE to the system is V=\(\frac {q}{a} – \frac {q}{2a} + \frac {q}{3a} – \frac {q}{4a} + \frac {q}{5a}\)……(up to infinity), where a, 2a, 3a… are the distances of the charges from the origin. Now, taking \(\frac {q}{a}\) common from the terms, we get V=\(\frac {q}{a} (1 – \frac {1}{2} + \frac {1}{3} – \frac {1}{4} + \frac {1}{5})\). By substituting the expansion of log2, we can get V=\(\frac {(q \, log2)}{a}\).

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