Thank you for reporting, we will resolve it shortly
Q.
If $q_{1}+q_{2}=q$, then the value of the ratio $\frac{q_{1}}{q}$, for which the force between $q_{1}$ and $q_{2}$ is maximum is
Electric Charges and Fields
Solution:
Let $r$ be the distance between $q_{1}$ and $q_{2}$.
According to Coulomb's law, the force between them is
$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
$=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}\left(q-q_{1}\right)}{r^{2}} $
$\left(\because q_{1}+q_{2}=q\right)$
For $F$ to be maximum $\frac{d F}{d q_{1}}=0 $
$\therefore \frac{d}{d q_{1}}\left[\frac{1}{4 \pi \varepsilon_{0} r^{2}}\left(q_{1} q-q_{1}^{2}\right)\right]=0$
$\Rightarrow q-2 q_{1}=0 $
$\Rightarrow \frac{q_{1}}{q}=\frac{1}{2}=0.5$
$\left[\text { As } \frac{1}{4 \pi \varepsilon_{0} r^{2}} \neq 0\right]$