Question

a charge of 120 microcoulomb is to be divided into two parts such that when kept at a certain distance the force between them may be maximum how should we divide

Answer 1

total charge Q is 120 microcoulomb F =
$k \times \frac{q(qtotal - q)}{ {r}^{2} }$
when F is maximum
$\frac{dy}{dq} = 0$
$\frac{d(qtotal \times q - {q}^{2}) }{dq}$
Q-2q = 0
q =
$\frac{qtotal}{2}$
q = 60 ×
${10}^{ - 6} coloumb$