Question

Three charges +q, Q and +q are placed in a straight line of length $l$ at 0, $\frac{l}{2}$ and $l$ respectively. What should be $Q$ in order to make the net force on +q to be zero?

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

Answer 1

Answer: (D)

-$\frac{q}{4}$

Answer 2

To find the value of $Q$ that makes the net force on $+q$ zero, we need to consider the electrostatic forces between the charges. Let's denote:

• $+q$ at $x = 0$
• $Q$ at $x = \frac{l}{2}$
• $+q$ at $x = l$

The force on the charge $+q$ at $x = 0$ is due to the other two charges. For the net force to be zero, the forces must balance each other.

1. Force due to $Q$ at $x = \frac{l}{2}$:

   The distance is $\frac{l}{2}$. The force is given by Coulomb's Law:

   $$F_1 = k \frac{|q \cdot Q|}{(\frac{l}{2})^2} = \frac{4k|qQ|}{l^2}$$

   where $k$ is Coulomb's constant. If $Q$ is negative, this force is attractive (towards $Q$), pulling $+q$ to the right.

2. Force due to $+q$ at $x = l$:

   The distance is $l$. The force is:

   $$F_2 = k \frac{q^2}{l^2}$$

   This force is repulsive, pushing $+q$ to the left.

For the net force to be zero, $|F_1| = |F_2|$:

$$\frac{4k|qQ|}{l^2} = \frac{kq^2}{l^2}$$

Simplifying:

$$4|Q| = q$$ $$|Q| = \frac{q}{4}$$

Since $Q$ must be negative to attract the $+q$ charge and balance the repulsive force from the other $+q$, we have:

$$Q = -\frac{q}{4}$$

Therefore, the value of $Q$ should be $-\frac{q}{4}$ to make the net force on $+q$ zero.