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# A charg Q is divided i  two parts such that when these two parts are kept at some sepration the force between them is maximum .find charge on each other

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

We get this with the help of some calculus :

F = K*[q1*q2]/r², q1+q2=Q

To maximize F(q1,q2) is to maximize q1*q2 so dF/dq=0

rewriting :

F(q1) = K*[q1*(Q-q1)] / r² => F(q1) = K*[q1Q-q1²)] / r²

dF(q1)/dq1 = [K/r²]* [Q-2q1] = 0 so Q- 2q1 = 0 .: q1 = Q/2 and q2=q1.

Remember and keep this result : the maximum A*B you can get is when A=B.

Any field.