Problem

A charge $q$ is placed at the center of the line joining two charges, each of magnitude $Q$. Prove that the system of three charges will be in equilibrium if $q = -\frac{Q}{4}$.

Data

• Separation between $Q$ and $q$: $r$
• Separation between $Q$ and $Q$: $2r$

Prerequisite Concepts

• Coulomb’s Law: The force between two charges is given by:

$$F = \frac{k q_1 q_2}{r^2}$$

Answer

Step 1: Equilibrium Condition

For equilibrium, the net force on $q$ due to the two charges $Q$ must be zero:

$$F_{(qQ)} + F_{(QQ)} = 0$$

Step 2: Force on $q$ due to $Q$ at a distance $r$

$$F_{(qQ)} = \frac{k q Q}{r^2}$$

Step 3: Force between $Q$ and $Q$ at a distance $2r$

$$F_{(QQ)} = \frac{k Q^2}{(2r)^2}$$

Simplify:

$$F_{(QQ)} = \frac{k Q^2}{4r^2}$$

Step 4: Net Force

For equilibrium:

$$\frac{k q Q}{r^2} + \frac{k Q^2}{4r^2} = 0$$

Factor $\frac{k Q}{r^2}$:

$$q + \frac{Q}{4} = 0$$

Step 5: Solve for $q$

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

Final Answer:
The system is in equilibrium if $q = -\frac{Q}{4}$.

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