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pallavi pradeep bhardwaj Grade: 12
        

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

7 years ago

Answers : (5)

Badiuddin askIITians.ismu Expert
147 Points
										

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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Askiitians Experts
Badiuddin

7 years ago
Karthik Pasupulatei
38 Points
										
i cant understand how todifrentiate f w.r.t q
 
one year ago
Hermione Weasley
12 Points
										
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
one year ago
Abhi
21 Points
										I have a doubt why df | da is done fgdswtyiplhxzssd hdgigc bhgdriov. My marks the correct the same problem for your email address for typos of you have received your mail to postmaster of u want it on a regular member and it will not work in progress of u vivek of my first thought it would have a look on your site at a very nice and warm weather was just
										
2 months ago
Ashish Kumar
36 Points
										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....
										
6 days ago
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