Question

A given charge \( Q \) is divided into two parts which are then kept at a distance \( d \) apart. The electrostatic force between them will be maximum if the two parts are

• A. \( \frac{Q}{4} \) and \( \frac{3Q}{4} \)
• B. \( \frac{7Q}{8} \) and \( \frac{Q}{8} \)
• C. \( \frac{Q}{3} \) and \( \frac{2Q}{3} \)
• D. \( \frac{5Q}{6} \) and \( \frac{Q}{6} \)
• E. \( \frac{Q}{2} \) each

Hint:

Product \( x(Q-x) \) is maximum when both are equal.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This is an optimization problem in electrostatics. We need to find how to divide a total charge Q into two parts, \(q_1\) and \(q_2\), such that the electrostatic force between them is maximized, keeping the distance of separation constant.
Step 2: Key Formula or Approach:
1. According to Coulomb's Law, the force between two charges \(q_1\) and \(q_2\) separated by a distance `d` is \(F = k \frac{q_1 q_2}{d^2}\), where `k` is Coulomb's constant. 2. Let the two parts be \(q\) and \(Q-q\), so that \(q_1 = q\) and \(q_2 = Q-q\). 3. The force is \(F(q) = k \frac{q(Q-q)}{d^2}\). 4. To maximize the force `F` with respect to `q`, we need to find the value of `q` for which the derivative \(\frac{dF}{dq}\) is zero.
Step 3: Detailed Explanation:
The force as a function of `q` is: \[ F(q) = \frac{k}{d^2} (qQ - q^2) \] Since `k` and `d` are constants, we only need to maximize the term \(f(q) = qQ - q^2\). Find the derivative of \(F(q)\) with respect to `q`: \[ \frac{dF}{dq} = \frac{k}{d^2} \frac{d}{dq}(qQ - q^2) \] \[ \frac{dF}{dq} = \frac{k}{d^2} (Q - 2q) \] Set the derivative to zero to find the maximum: \[ \frac{k}{d^2} (Q - 2q) = 0 \] Since \(\frac{k}{d^2} \neq 0\), we must have: \[ Q - 2q = 0 \] \[ 2q = Q \implies q = \frac{Q}{2} \] So, the first part is \(q_1 = q = \frac{Q}{2}\). The second part is \(q_2 = Q - q = Q - \frac{Q}{2} = \frac{Q}{2}\). The force is maximum when the charge is divided into two equal halves.
Step 4: Final Answer:
The two parts should be \(\frac{Q}{2}\) each.