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A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium of q is equal to (a) -(Q/2) (b) -(Q/4) (c) + (Q/4) (d) +(Q/2)


A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges


will be in equilibrium of q is equal to


(a) -(Q/2) (b) -(Q/4) (c) + (Q/4) (d) +(Q/2)


 


Grade:11

2 Answers

Anirban Mukherjee
47 Points
9 years ago

Here use this formula

U = 1/(4Πε0)[2qQ/r + Q2/2r] = 0

q=-Q/4

Jevin George
14 Points
9 years ago

well...you hv not mentioned what q is in ur ques....

,,solving ques assuming that q is the third charge that u mentioned

for the first Q charge to be in equilibrium...net force on it shud be zero..

i.e... force due to q + force due to Q =0

let the distance btw Q charge to Q charge be 2l.....so dat distance btw q and Q is l

kqQ/l2+kQ2/4l2=0

solving it we get..q=-Q/4...

..hope it helps..:)

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