Question

A charge \( q \) is placed at the centre of the line joining two equal charges \( Q \). The system of the three charges will be in equilibrium if \( q \) is equal to:

• A. \( -\frac{Q}{2} \)
• B. \( -\frac{Q}{4} \)
• C. \( \frac{Q}{2} \)
• D. \( \frac{Q}{4} \)

Hint:

To achieve equilibrium in a system of charges, the net force on each charge must be zero. Use Coulomb's law to calculate the forces.

Answer 1

The Correct Option is A

A system of three charges is in equilibrium when the net force on each charge is zero. Let us consider two equal charges \( Q \) placed at the positions \( -a \) and \( a \) on the x-axis, with the charge \( q \) placed at the origin.

For charge \( q \) to be in equilibrium, the net force acting on it due to charges \( Q \) needs to be zero. The forces exerted on \( q \) by \( Q \) are equal in magnitude and opposite in direction. If \( F_1 \) is the force due to the charge at \( -a \) and \( F_2 \) is the force due to the charge at \( a \), then \( F_1 = F_2 \).

By Coulomb's Law, the force \( F \) between two point charges is given by:

\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]

Here, the forces can be represented as:

\[ F_1 = \frac{k \cdot |Q \cdot q|}{a^2} \]

\[ F_2 = \frac{k \cdot |Q \cdot q|}{a^2} \]

Since \( F_1 \) and \( F_2 \) are equal and acting oppositely, the charge \( q \) does not move.

For charge Q at -a and at a to be in equilibrium due to charge \( q \) an additional condition is needed. The forces on these charges Q must be balanced. The force \( F_q \) on charge at \( -a \) due to central charge \( q \) is:

\[ F_q = \frac{k \cdot |Q \cdot q|}{(2a)^2} \]

The force between charges \( Q \) at positions \( -a \) and \( a \) should be equal and opposite to F_q for equilibrium condition:

\[ F_Q = \frac{k \cdot Q^2}{(2a)^2} \]

Setting \( F_Q = F_q \):

\[ \frac{k \cdot Q^2}{(2a)^2} = \frac{k \cdot |Q \cdot q|}{a^2} \]

Solving for \( q \), we have:

\[ Q^2 = 4 \cdot Q \cdot q \]

\[ q = \frac{Q^2}{4Q} = \frac{Q}{4} \text{ but with opposite sign for equilibrium: } q = -\frac{Q}{2} \]

Thus, for the system to be in equilibrium, the charge \( q \) should be:

\(\boxed{-\frac{Q}{2}}\)

Answer 2

This question asks what central charge keeps three charges in a line balanced. Rather than deriving the value from scratch, each option can be tested directly against the condition needed for equilibrium.

1. \( -\frac{Q}{2} \): Placing the two outer charges \( Q \) at distance \( a \) from the centre, they repel each other with force \( \frac{kQ^2}{(2a)^2} \). The pull from a central charge of size \( \frac{Q}{2} \) on one outer charge is \( \frac{kQ(Q/2)}{a^2} = \frac{kQ^2}{2a^2} \). Comparing this to the repulsion \( \frac{kQ^2}{4a^2} \) shows the pull is twice as strong as needed, so this magnitude overcorrects.
2. \( -\frac{Q}{4} \): The pull from a central charge of size \( \frac{Q}{4} \) on one outer charge is \( \frac{kQ(Q/4)}{a^2} = \frac{kQ^2}{4a^2} \), which exactly matches the repulsion between the two outer charges, \( \frac{kQ^2}{(2a)^2} = \frac{kQ^2}{4a^2} \). The negative sign makes this pull attractive, directly opposing the repulsion, so this value balances the outer charge exactly.
3. \( \frac{Q}{2} \): Same size issue as the first option (double the needed pull), and on top of that, a positive sign here would repel the outer charge rather than pull it in, adding to the existing repulsion instead of cancelling it.
4. \( \frac{Q}{4} \): This has the right size but a positive sign would push the outer charge further away rather than pulling it back, working against equilibrium instead of creating it.

The centre charge only balances the system when it is both the right size, a quarter of \( Q \), and attractive toward the outer charges, meaning negative if \( Q \) is positive. Checking the force balance directly on one of the outer charges confirms this.

So the correct answer is \( -\frac{Q}{4} \).