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`        A charge Q is divided in two parts such that when these two parts are kept at some sepeation then the force between them is maximum.find charge on each other`
8 months ago

```							Dear Tanmay  let parts are sepereted at a distance of dso 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/d2 [Q -2q] for maximum force dF/dq=0                             Q-2q=0                              Q/q =2
```
8 months ago
```							We get this with the help of some calculus :F = K*[q1*q2]/r², q1+q2=QTo maximize F(q1,q2) is to maximize q1*q2 so dF/dq=0rewriting :F(q1) = K*[q1*(Q-q1)] / r² => F(q1) = K*[q1Q-q1²)] / r²dF(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.
```
8 months ago
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