A total charge `Q` is broken in two parts `Q_(1)` and `Q_(2)` and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

Options

A. `Q_(2) = (Q)/(R) , Q_(1) = Q = (Q)/(R)`
B. `Q_(2) = (Q)/(4) , Q_(1) = Q - (2 Q)/(3)`
C. `Q_(2) = (Q)/(4), Q_(1) = (3Q)/(3)`
D. `Q_(1) = (Q)/(2) , Q_(2) = (Q)/(2)`

Answer

D

Explanation

`Q_(1) + Q_(2) = Q . . . (i) and F = k (Q_(1) Q_(2))/(r^(2)) ` . . . (ii)
From (i) and (ii) ` F = (kQ_(1) (Q - Q_(1)))/( r^(2))`
For F to be maximum `(dF)/( d Q_(1)) = 0 rArr Q_(1) = Q_(2) = (Q)/(2)`