A charge q is divided into two parts q1 and (q-q1) .What is the ratio of q/q1 so that the force between the two parts placed at a given distance apart is maximum.
options are - a)1:1 b) 2:1 c)1:2 d)1:4

Question

A charge q is divided into two parts q1 and (q-q1) .What is the ratio of q/q1 so that the force between the two parts placed at a given distance apart is maximum.
options are - a)1:1   b) 2:1  c)1:2  d)1:4

anonymous · Answer

the way i did it was - 
F = kq1(q-q1)/ r^2;
this force has to be maximum ; 
so differentiation of maxima = 0;
therefore differentiation of F = 0;
 but i have no clue how to differentiate it..:(

anonymous · Answer

@mos1635 - help me again please..!

anonymous · Answer

The answer is 1:1

$$dF/dq1= 0$$  for maximum force.

$$dF/dq1= (k/r ^{2} ) * (q-2q1) = 0$$

Therefore
$$q-2q1=0$$
$$q _{1}$$

Therefore both charges are q/2

anonymous · Answer

q1=q/2

mos1635 · Answer

F=K/r^2*q1(q-q1)
F=K/r^2*(q*q1-q1^2)
dF/dq1=k/r^2*[d/dq1(q*q1-q1^2)]
dF/dq1=k/r^2*[d/dq1(q*q1)-d/dq1q1^2)]
dF/dq1=k/r^2*(q-2q1)
to be 0
q-2q1=0
q/q1=2

anonymous · Answer

Sorry the answer is q/q1=2   . I made a mistake in first line of my solution.  I misread the question

anonymous · Answer

thanks guys..:D