Q199 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Physics
A charge $Q$ is divided into two parts and placed at a given distance so that the force between them is maximum. The value of each charge is:
- Correct answerA$\dfrac{Q}{2}, \dfrac{Q}{2}$
- B$\dfrac{3Q}{4}, \dfrac{Q}{4}$
- C$\dfrac{Q}{3}, \dfrac{2Q}{3}$
- D$\dfrac{2Q}{3}, \dfrac{Q}{3}$
Solution
Let the two parts of the charge $Q$ be $q$ and $Q - q$.
The electrostatic force between them when placed at a given distance $r$ is given by Coulomb's law:
$F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q(Q - q)}{r^2}$
For the force to be maximum, the first derivative of $F$ with respect to $q$ must be zero:
$\dfrac{dF}{dq} = 0$
$\dfrac{1}{4\pi\varepsilon_0 r^2} \dfrac{d}{dq}(qQ - q^2) = 0$
Since $\dfrac{1}{4\pi\varepsilon_0 r^2}$ is a non-zero constant, we get:
$\dfrac{d}{dq}(qQ - q^2) = 0$
$Q - 2q = 0$
$2q = Q$
$q = \dfrac{Q}{2}$
The other part of the charge is $Q - q = Q - \dfrac{Q}{2} = \dfrac{Q}{2}$.
Thus, the value of each charge is $\dfrac{Q}{2}$ and $\dfrac{Q}{2}$.
The electrostatic force between them when placed at a given distance $r$ is given by Coulomb's law:
$F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q(Q - q)}{r^2}$
For the force to be maximum, the first derivative of $F$ with respect to $q$ must be zero:
$\dfrac{dF}{dq} = 0$
$\dfrac{1}{4\pi\varepsilon_0 r^2} \dfrac{d}{dq}(qQ - q^2) = 0$
Since $\dfrac{1}{4\pi\varepsilon_0 r^2}$ is a non-zero constant, we get:
$\dfrac{d}{dq}(qQ - q^2) = 0$
$Q - 2q = 0$
$2q = Q$
$q = \dfrac{Q}{2}$
The other part of the charge is $Q - q = Q - \dfrac{Q}{2} = \dfrac{Q}{2}$.
Thus, the value of each charge is $\dfrac{Q}{2}$ and $\dfrac{Q}{2}$.
Hence, the correct answer is option: 'A'.