Question:
A charge \(Q\) is to be divided between two objects. The values of the charges on the objects so that the electrostatic force between them will be maximum is:
A charge \(Q\) is to be divided between two objects. The values of the charges on the objects so that the electrostatic force between them will be maximum is:
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For a fixed sum, the product of two quantities is maximum when the two quantities are equal.
Updated On: Jun 24, 2026
- \(\dfrac{Q}{2},\dfrac{Q}{2}\)
- \(\dfrac{Q}{3},\dfrac{2Q}{3}\)
- \(\dfrac{Q}{4},\dfrac{3Q}{4}\)
- \(\dfrac{Q}{5},\dfrac{4Q}{5}\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the electrostatic force formula.
According to Coulomb’s law, \[ F=\frac{kq_1q_2}{r^2} \] Since \[ k \] and \[ r \] are constants, the force depends on \[ q_1q_2 \]
Step 2: Use the condition for total charge.
Let the charges be \[ q \] and \[ Q-q \] Then, \[ F\propto q(Q-q) \] \[ F\propto qQ-q^2 \]
Step 3: Maximize the expression.
The expression \[ qQ-q^2 \] is a quadratic function.
Differentiate with respect to \(q\): \[ \frac{d}{dq}(qQ-q^2)=Q-2q \] For maximum force, \[ Q-2q=0 \] \[ q=\frac{Q}{2} \] Thus, \[ Q-q=\frac{Q}{2} \]
Step 4: Final conclusion.
Hence, the electrostatic force becomes maximum when the charges are \[ \boxed{\frac{Q}{2},\frac{Q}{2}} \]
According to Coulomb’s law, \[ F=\frac{kq_1q_2}{r^2} \] Since \[ k \] and \[ r \] are constants, the force depends on \[ q_1q_2 \]
Step 2: Use the condition for total charge.
Let the charges be \[ q \] and \[ Q-q \] Then, \[ F\propto q(Q-q) \] \[ F\propto qQ-q^2 \]
Step 3: Maximize the expression.
The expression \[ qQ-q^2 \] is a quadratic function.
Differentiate with respect to \(q\): \[ \frac{d}{dq}(qQ-q^2)=Q-2q \] For maximum force, \[ Q-2q=0 \] \[ q=\frac{Q}{2} \] Thus, \[ Q-q=\frac{Q}{2} \]
Step 4: Final conclusion.
Hence, the electrostatic force becomes maximum when the charges are \[ \boxed{\frac{Q}{2},\frac{Q}{2}} \]
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