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A charge Q is divide intointo parts and then they are placed at a fixed. The force between the two charges is always maximum when charges are

2 years ago

let parts are sepereted at a distance of d

so repulsion force F= k q(Q-q)/d2

here Q and d are constant so for maximum force differentiate the force w.r.t q

dF/dq = k/d[Q -2q]

for maximum force dF/dq=0

Q-2q=0

Q/q =2

2 years ago
Dear student
Let q1 = qand q2 = Q-q
r  be the distance or separation between them
The force of repusion between them is
F = k (Q-q)q/r^2
Differenciating F with respect to q and setting it to zero will give us extreme force.
dF/dq = 0
so, q/Q = 1/2
For this value q/Q the force is maximum .
The force will be maximum if the second derivative of F is less than 0
d^2F /dq^2= -2k/r^2 which is less than 0.
Thus the force of repulsion is maximum when q=Q/2
so , q1 = Q/2 an q2 = Q/2

Hope this helps