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# A charge Q is divided into two parts of qand Q-q if the coulomb repulsion between them when they are seprated is to be maximum the ratio of Q/q should be Grade:12

## 10 Answers Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear pallavi

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/d2 [Q -2q]

for maximum force dF/dq=0

Q-2q=0

Q/q =2

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Badiuddin

5 years ago
F = kq(Q-q)/r^2
=> F = k(Qq-q^2)/r^2
Since Q, r, and k are all constants,
Differentiating,
dF/dq = d (k(Qq-q^2)/r^2) / dq
=> dF/dq = k/r^2 * [Q – 2q]
Since the maximum force is a constant, dF/dq must be equal to 0.
Hence, k/r^2 * [Q – 2q] = 0
=> Q – 2q = 0
=> Q = 2q
=> q = Q/2 or Q/q = 2
4 years ago
Let q and Q-q be the charges on the two objects. Then force between the two object is F =1/ 4 pi Eò . q(Q-q)/ r^2 where r is the distance between them....For F to be max , dF/dq = 0 Or 1/4pi E0 .r^2 .d(qQ-q^2)/ dq= 0 Q-2q =0 q = Q/2 answer....
3 years ago
We equate dF/dq=0 because it`s one of the rules of differentiation where we equate it to 0 when we want any maximum or minimum value of something
3 years ago
q and (Q-q) be charges
So the force (F) =Kq(Q-q)/r2
To maximize dF/dq=0
so (q-Q+q=0)
q=Q/2
in other words, two charge force will increase as the multiple will increase, but when the multiple will be maximum at the mid point only.
2 years ago
Differentiating gives us the local maxima(maximum value) and minima(minimum value) of a curve. So I guess that's why we differentiate.
one year ago
Hello Pallavi,
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/d2 [Q -2q]
for maximum force dF/dq=0
Q-2q=0
Q/q =2
Thanks
I hope above solution will clear your all doubts.
Please feel free to post and ask as much doubts as possible.
All the best.

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