Three charges -q, Q and -q are placed
respectively at equal distances on a straight line.
If the potential energy of the system of three
charges is zero, then what is the ratio of Q:q?
Answers
Explanation:
i have considered the distance between two charges to be R. After writing the potential energy of the system we get the ratio as 1:4.
ratio of charge Q : q = 1 : 4.
let distance between -q and Q is r.
a/c to question, -q , Q and -q are placed respectively at equal distance on a straight line.
-q .......r....... Q .......r......-q
total potential energy of system = potential energy of -q and Q + potential energy of Q and -q + potential energy of -q and -q
⇒0 = k(-q)Q/r + kQ(-q)/r + k(-q)(-q)/2r
⇒0 = -kqQ/r - kqQ/r + kq²/2r
⇒0 = -2Q/1 + q/2
⇒2Q = q/2
⇒Q/q = 1/4
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