# Question #eb655

Oct 28, 2016

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

$F = k \frac{q \left(Q - q\right)}{r} ^ 2. \ldots . \left(1\right)$

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.

1. Algebraic process to determine the condition for the maximum value of F

Now

$F = k \frac{q Q - {q}^{2}}{r} ^ 2$

$\implies F = \frac{k}{r} ^ 2 \left({Q}^{2} / 4 - {Q}^{2} / 4 + 2 \cdot \frac{Q}{2} \cdot q - {q}^{2}\right)$

$\implies F = \frac{k}{r} ^ 2 \left({Q}^{2} / 4 - \left({\left(\frac{Q}{2}\right)}^{2} - 2 \cdot \frac{Q}{2} \cdot q + {q}^{2}\right)\right)$

$\implies F = \frac{k}{r} ^ 2 \left({Q}^{2} / 4 - {\left(\frac{Q}{2} - q\right)}^{2}\right)$

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

$\frac{Q}{2} - q = 0$
$\implies q = \frac{Q}{2}$

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

$\frac{Q}{q} = \frac{2}{1}$

1. Calculus process

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

$\frac{\mathrm{dF}}{\mathrm{dq}} = \frac{k}{r} ^ 2 \left(\frac{d}{\mathrm{dq}} \left(Q q - {q}^{2}\right)\right)$

$\implies \frac{\mathrm{dF}}{\mathrm{dq}} = \frac{k}{r} ^ 2 \left(Q \frac{d}{\mathrm{dq}} \left(q\right) - \frac{d}{\mathrm{dq}} \left({q}^{2}\right)\right)$

$\implies \frac{\mathrm{dF}}{\mathrm{dq}} = \frac{k}{r} ^ 2 \left(Q - 2 q\right)$

For maximum value of F

$\frac{\mathrm{dF}}{\mathrm{dq}} = 0$

This means $\frac{k}{r} ^ 2 \left(Q - 2 q\right) = 0$

$\implies Q = 2 q$

$\implies \frac{Q}{q} = \frac{2}{1}$