Question

Let a charge Q on a body be divided into two charged bodies and these bodies thereafter are kept at certain distance from each other. Show that the force between the bodies is maximum if Q/q = 2.

Answer 1

Correct Question:-

Let a charge $\sf{Q}$ on a body be divided into two charges, $\sf{q}$ and $\sf{Q-q,}$ and these bodies thereafter are kept at certain distance $\sf{r}$ from each other. Show that the force $\sf{F}$ between these bodies is maximum if $\sf{\dfrac{Q}{q}=2}.$

Solution:-

Force acting between two charged particles $\sf{q_1}$ and $\sf{q_2}$ separated by a distance $\sf{r}$ is given by,

$\sf{\longrightarrow F=\dfrac{kq_1q_2}{r^2}}$

where $\sf{k=\dfrac{1}{4\pi\epsilon_0}}$ is a constant.

In the question the charges are $\sf{q}$ and $\sf{Q-q.}$

Hence the force,

$\sf{\longrightarrow F=\dfrac{kq(Q-q)}{r^2}}$

If this force is maximum, its derivative with respect to $\sf{q}$ should be zero.

$\sf{\longrightarrow \dfrac{dF}{dq}=0}$

$\sf{\longrightarrow \dfrac{d}{dq}\left[\dfrac{kq(Q-q)}{r^2}\right]=0}$

$\sf{\longrightarrow \dfrac{k}{r^2}\cdot\dfrac{d}{dq}\big[q(Q-q)\big]=0}$

Since $\sf{\dfrac{k}{r^2}}$ is non - zero,

$\sf{\longrightarrow\dfrac{d}{dq}\big[q(Q-q)\big]=0}$

By product rule, ($\sf{Q}$ is a constant charge)

$\sf{\longrightarrow 1(Q-q)+q(0-1)=0}$

$\sf{\longrightarrow Q-q-q=0}$

$\sf{\longrightarrow Q-2q=0}$

$\sf{\longrightarrow Q=2q}$

$\sf{\longrightarrow\underline{\underline{\dfrac{Q}{q}=2}}}$

Hence the Proof!