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Question

Physics Question on Electric charges and fields

A charge QQ is to be divided into two parts, qq and (Qq)(Q- q), such that force between them is maximum. Then

A

Q=3qQ = 3q

B

Q=1.5qQ = 1.5 q

C

Q=2qQ = 2q

D

Q=4qQ = 4 q

Answer

Q=2qQ = 2q

Explanation

Solution

F=kq(Qq)r2F =k \frac{q\left(Q - q\right)}{r^{2}}
dFdq=kr2[(Qq)+q(1)]\frac{dF}{dq} = \frac{k}{r^{2}} \left[\left(Q -q\right) + q\left(-1\right)\right]
=kr2(Q2q)\, \, \, \, = \frac{k}{r^{2}} \left(Q -2q\right)
For maximum value of F,dFdq=0F, \frac{dF}{dq} = 0
kr2(Q2q)=0\therefore \, \frac{k}{r^{2} } \left(Q - 2q\right)= 0
or, Q=2qQ = 2q