Two particles (free to move) with charges `+q` and `+4q` are a distance L apart. A third charge is placed so that the entire system is in equilibrium.
(a) Find the location, magnitude and sign of the third charge.
(b) Show that the equilibrium is unstable.

Options

A. `Q=(4)/(9)q "(negative)at" (l)/(3)`
B. `Q=(4)/(9) q"(positive) at" (l)/(3)`
C. `Q=q "(positive) at" (l)/(3)`
D. `Q = q"(negative) at" (l)/(3)`

Answer

To solve the problem step-by-step, we will break it down into parts (a) and (b) as specified in the question.

### Part (a): Finding the location, magnitude, and sign of the third charge

1. **Understanding the System**: We have two charges, \( +q \) at point A and \( +4q \) at point B, separated by a distance \( L \). We need to place a third charge \( Q \) such that the entire system is in equilibrium.

2. **Assuming the Position of Charge \( Q \)**: Let’s assume the charge \( Q \) is placed at a distance \( x \) from charge \( +q \). Consequently, the distance from charge \( +4q \) will be \( L - x \).

3. **Setting Up the Forces**: For the system to be in equilibrium, the net force acting on charge \( Q \) must be zero. The force exerted on \( Q \) by \( +q \) (let's call it \( F_1 \)) must be equal to the force exerted on \( Q \) by \( +4q \) (let's call it \( F_2 \)). Since both charges are positive, the force \( F_1 \) will be repulsive and \( F_2 \) will be attractive if \( Q \) is negative.

\[ F_1 = k \frac{q |Q|}{x^2} \]
\[ F_2 = k \frac{4q |Q|}{(L - x)^2} \]

4. **Equating the Forces**: For equilibrium, we set \( F_1 = F_2 \):

\[ k \frac{q |Q|}{x^2} = k \frac{4q |Q|}{(L - x)^2} \]

We can cancel \( k \) and \( q \) (assuming \( Q \neq 0 \)):

\[ \frac{|Q|}{x^2} = \frac{4 |Q|}{(L - x)^2} \]

5. **Simplifying the Equation**: This simplifies to:

\[ (L - x)^2 = 4x^2 \]

Taking the square root:

\[ L - x = 2x \quad \text{or} \quad L - x = -2x \]

6. **Solving for \( x \)**: From \( L - x = 2x \):

\[ L = 3x \implies x = \frac{L}{3} \]

From \( L - x = -2x \):

\[ L = -x \implies x = -L \quad \text{(not valid)} \]

Thus, the valid solution is \( x = \frac{L}{3} \).

7. **Finding the Magnitude and Sign of Charge \( Q \)**: Now we need to find the magnitude of \( Q \). We can use the equilibrium condition on charge \( +4q \):

\[ F_3 = F_4 \]

Where \( F_3 \) is the force on \( +4q \) due to \( Q \) and \( F_4 \) is the force on \( +4q \) due to \( +q \):

\[ k \frac{4q |Q|}{(L - x)^2} = k \frac{q \cdot 4q}{L^2} \]

Substituting \( x = \frac{L}{3} \):

\[ k \frac{4q |Q|}{\left( \frac{2L}{3} \right)^2} = k \frac{4q^2}{L^2} \]

Simplifying gives:

\[ |Q| = \frac{q \cdot \frac{4L^2}{9}}{4 \cdot \frac{L^2}{9}} = -\frac{4q}{9} \]

Therefore, the charge \( Q \) is \( -\frac{4q}{9} \).

### Summary of Part (a):
- Location of charge \( Q \): \( \frac{L}{3} \) from charge \( +q \)
- Magnitude and sign of charge \( Q \): \( -\frac{4q}{9} \)

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### Part (b): Showing that the equilibrium is unstable

1. **Displacing Charge \( Q \)**: To show that the equilibrium is unstable, we will consider a small displacement \( d \) of charge \( Q \) to the right.

2. **New Distances**: After displacement, the distance from \( +q \) becomes \( \frac{L}{3} + d \) and the distance from \( +4q \) becomes \( \frac{2L}{3} - d \).

3. **Calculating Forces After Displacement**:
- The force \( F_1 \) (from \( +q \)) will decrease because the distance has increased.
- The force \( F_2 \) (from \( +4q \)) will increase because the distance has decreased.

Thus, we have:

\[ F_1' = k \frac{q \cdot \left| -\frac{4q}{9} \right|}{\left( \frac{L}{3} + d \right)^2} \]
\[ F_2' = k \frac{4q \cdot \left| -\frac{4q}{9} \right|}{\left( \frac{2L}{3} - d \right)^2} \]

4. **Net Force Direction**: Since \( F_1' \) decreases and \( F_2' \) increases, the net force acting on \( Q \) will be directed away from the equilibrium position. This means that if \( Q \) is displaced slightly, it will experience a net force pushing it further away from equilibrium.

### Conclusion for Part (b):
- The equilibrium is unstable because any small displacement of charge \( Q \) results in a net force that pushes it further away from the equilibrium position.

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