Question

Charge $Q$, is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are having the charge.

• A. $\frac{Q}{2}$ each
• B. $\frac{Q}{4}$ and $\frac{3Q}{4}$
• C. $\frac{Q}{3}$ and $\frac{2Q}{3}$
• D. $e$ and $(Q-e)$, where $e=$ electronic charge

Answer 1

Answer: (A)

$\text{Step 1: Calculation of force}$

Let one part of the charge be $q$. Thus other part of the charge is $Q-q$

From Coulomb's law:

Electrostatic force between them $F=\frac{kq(Q-q)}{r^2}$

$\text{Step 2: Condition of maximum force}$

For maximum force, $\frac{dF}{dq}=0$

$\therefore$ $\frac{d}{dq}(\frac{kq(Q-q)}{r^2})=0$

Or, $\frac{d}{dq}(qQ-q^2)=0$

Or, $Q-2q=0$

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

Hence force between them is maximum if the two parts have equal charge $Q/2$ each. So, Option A is correct