Question

Given that $q_1+q_2=q$. If the force between $q_1$ and $q_2$ is maximum, then $\frac{q_1}{q_2}$ is

• A. $1$
• B. $0.75$
• C. $0.5$
• D. $0.25$

Answer 1

Answer: (A)

$1$

$F=\frac{K{q}_1{q}_2}{r^2}$

$⟹F=\frac{K{q}_1(q-q_1)}{r^2}$

Differentiating both sides with respect to $q_1-$

$\frac{K}{r^2}(q-2q_1)=\frac{dF}{d{q}_1}$

For, force to be maximum, $\frac{dF}{d{q}_1}=0$

$⟹q-2q_1=0$

$⟹q_1=\frac{q}{2}$

$q_2=q-q_1=\frac{q}{2}$

Hence, $\frac{q_1}{q_2}=1$

Answer-(A)