Question

A point charge $Q$ is placed at the centre of the line joining two equal point charges $+q$ and $+q$. The value of $Q$ if the system of the charges is in equilibrium, is

• A. $-\frac{q}{2}$
• B. $-\frac{q}{4}$
• C. $+\frac{q}{4}$
• D. $+\frac{q}{2}$

Hint:

You can determine the sign of $Q$ using basic logic! Since the two outer charges are both positive ($+q$), they naturally repel each other outward. To keep them from flying apart, the central charge $Q$ must be negative to exert an attractive inward force. This immediately eliminates options (C) and (D)!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Three charges lie along a straight line. Two identical charges $+q$ are fixed at a distance $r$ from each other, and a third charge $Q$ is placed exactly midway between them. We need to find the value and sign of $Q$ such that the entire system remains in a stable mechanical equilibrium.

Step 2: Key Formula or Approach:
For the system to be in equilibrium, the net electrostatic force acting on any individual charge must be exactly zero. 1. The central charge $Q$ is automatically in equilibrium because it experiences equal and opposite forces from the two identical $+q$ charges on either side. 2. Therefore, we look at the equilibrium condition of one of the outer $+q$ charges. The vector sum of forces from the other $+q$ charge and the central charge $Q$ must equal zero: $$\sum F = F_q + F_Q = 0$$ 3. Use Coulomb's Law: $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{d^2}$.

Step 3: Detailed Explanation:
Let the two $+q$ charges be placed at a distance $r$ apart. The central charge $Q$ sits at a distance of $\frac{r}{2}$ from each charge. Let's write the net force acting on the $+q$ charge at the far right: Force due to the other $+q$ charge at distance $r$: $$F_q = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2}$$ Force due to the central charge $Q$ at distance $\frac{r}{2}$: $$F_Q = \frac{1}{4\pi\varepsilon_0} \frac{qQ}{\left(\frac{r}{2}\right)^2} = \frac{1}{4\pi\varepsilon_0} \frac{4qQ}{r^2}$$ For this outer charge to remain stationary, the sum of these forces must be zero: $$\frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2} + \frac{1}{4\pi\varepsilon_0} \frac{4qQ}{r^2} = 0$$ Cancel out the common constant terms $\frac{1}{4\pi\varepsilon_0}$ and $\frac{q}{r^2}$: $$q + 4Q = 0$$ Isolating $Q$: $$4Q = -q \implies Q = -\frac{q}{4}$$ The negative sign indicates that the central charge must have an opposite polarity to pull the outer charges inward, counteracting their mutual repulsion.

Step 4: Final Answer:
The value of the charge $Q$ is $-\frac{q}{4}$, which corresponds to option (B).