Question #eb655

Question

1985 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-28T08:05:55

By coulomb's law the force acting between two point charges q and Q-q lying r distance apart is given by

#F=k(q(Q-q))/r^2.....(1)#

Where Q is the total charge, a constant quantity.
k is the Coulmb's constant

q is the varible charge on which the magnitude of force F depends.

Now

#F=k(qQ-q^2)/r^2#

#=>F=k/r^2(Q^2/4-Q^2/4+2*Q/2*q-q^2)#

#=>F=k/r^2(Q^2/4-((Q/2)^2-2*Q/2*q+q^2))#

#=>F=k/r^2(Q^2/4-(Q/2-q)^2)#

This relation suggests that the magnltude of force F will be maximum only when

#Q/2-q=0#
#=>q=Q/2#

This means the force between the two parts of Q will be maximum only when Q is divided in two equal parts and then

#Q/q=2/1#

Differentiating equation (1) w .r to q we get

#(dF)/(dq)=k/r^2(d/(dq)(Qq-q^2))#

#=>(dF)/(dq)=k/r^2(Qd/(dq)(q)-d/(dq)(q^2))#

#=>(dF)/(dq)=k/r^2(Q-2q)#

For maximum value of F

#(dF)/(dq)=0#

This means #k/r^2(Q-2q)=0#

#=>Q=2q#

#=>Q/q=2/1#