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.

Explanation

To solve the problem step by step, we will approach each part of the question systematically.

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

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

2. **Position of the Third Charge**: Let’s denote the position of the third charge \( Q \) as being at a distance \( x \) from the charge \( +q \). Consequently, the distance from the charge \( +4q \) will be \( L - x \).

3. **Forces Acting on Charge \( Q \)**:
- The force \( F_1 \) on charge \( Q \) due to \( +4q \) is attractive (since \( Q \) will be negative) and can be expressed as:
\[ F_1 = k \frac{4q \cdot |Q|}{(L - x)^2} \]
- The force \( F_2 \) on charge \( Q \) due to \( +q \) is also attractive and can be expressed as:
\[ F_2 = k \frac{q \cdot |Q|}{x^2} \]

4. **Setting Up the Equilibrium Condition**: For the system to be in equilibrium, the net force on charge \( Q \) must be zero:
\[ F_1 = F_2 \]
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ k \frac{4q \cdot |Q|}{(L - x)^2} = k \frac{q \cdot |Q|}{x^2} \]
Canceling \( k \) and \( |Q| \) (assuming \( |Q| \neq 0 \)):
\[ \frac{4q}{(L - x)^2} = \frac{q}{x^2} \]
Simplifying gives:
\[ 4x^2 = (L - x)^2 \]

5. **Solving the Equation**:
Expanding and rearranging:
\[ 4x^2 = L^2 - 2Lx + x^2 \]
\[ 3x^2 + 2Lx - L^2 = 0 \]
This is a quadratic equation in \( x \). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2L \pm \sqrt{(2L)^2 - 4 \cdot 3 \cdot (-L^2)}}{2 \cdot 3} \]
\[ x = \frac{-2L \pm \sqrt{4L^2 + 12L^2}}{6} = \frac{-2L \pm \sqrt{16L^2}}{6} = \frac{-2L \pm 4L}{6} \]
This gives two solutions:
\[ x = \frac{2L}{6} = \frac{L}{3} \quad \text{(valid)} \quad \text{and} \quad x = \frac{-6L}{6} = -L \quad \text{(not valid)} \]

6. **Magnitude and Sign of Charge \( Q \)**:
Now we need to find the magnitude of charge \( Q \). We can use the equilibrium condition for charge \( +q \):
\[ F_1' = F_2' \]
Where:
\[ F_1' = k \frac{4q \cdot q}{L^2}, \quad F_2' = k \frac{q \cdot |Q|}{(L/3)^2} \]
Setting these equal gives:
\[ k \frac{4q^2}{L^2} = k \frac{q \cdot |Q|}{(L/3)^2} \]
Canceling \( k \) and \( q \) (assuming \( q \neq 0 \)):
\[ 4 = |Q| \cdot \frac{9}{L^2} \]
Thus:
\[ |Q| = \frac{4L^2}{9} \]
Since \( Q \) must be negative to attract both positive charges:
\[ Q = -\frac{4q}{9} \]

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

1. **Understanding Stability**: For equilibrium to be stable, if the system is slightly disturbed, it should return to its original position. If it moves away, it is unstable.

2. **Analyzing Forces**: If the charge \( Q \) is slightly moved towards \( +q \) or \( +4q \), the forces acting on \( +q \) and \( +4q \) will change. Both charges will repel each other more strongly due to their positive nature.

3. **Conclusion**: Since both \( +q \) and \( +4q \) will repel each other, if \( Q \) is disturbed, the forces will not restore it to equilibrium but rather push it further away. Thus, the equilibrium is unstable.

### Final Answer
(a) The third charge \( Q \) is located at \( \frac{L}{3} \) from \( +q \), has a magnitude of \( -\frac{4q}{9} \), and is negative.
(b) The equilibrium is unstable because any small displacement of charge \( Q \) will lead to forces that push it further away from its equilibrium position.