A charge \( Q \) is divided into two parts \( q \) and \( (Q-q) \). If the force of repulsion for a fixed separation between them is to be maximum, then, the ratio of \( Q \) and \( q \) should be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2:1 \)
By Coulomb’s law, the force between the two parts is proportional to the product of their charges, since the separation r is fixed.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q(Q-q)}{r^{2}} \)
For the maximum force, set the derivative with respect to q equal to zero:
\( \displaystyle \frac{dF}{dq} \propto Q - 2q = 0 \)
\( \displaystyle q = \frac{Q}{2} \)
\( \displaystyle \frac{Q}{q} = \frac{2}{1} \)
So the ratio \( Q:q = 2:1 \).
Shortcut: The sum of the two parts is fixed at Q, and the product of two numbers with a fixed sum is greatest when they are equal. So \( q = Q - q \), giving \( q = Q/2 \).
Note: The same question was also asked in CUET (UG) 2023 on 23-05-2023 (Slot 3).