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:

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