Two charges `-q` each are separated by distance `2d`. A third charge `+q` is kept at mid-point O. Find potential energy of `+q` as a function of small distance `x` from O due to `-q` charges. Prove that the charge at O is in an unstable equilibrium.

Answer

To solve the problem, we need to find the potential energy of the charge \( +q \) at the midpoint \( O \) due to the two charges \( -q \) that are separated by a distance \( 2d \). We will then show that this configuration represents an unstable equilibrium.

### Step 1: Define the Configuration
We have two charges \( -q \) located at points \( A \) and \( B \), such that the distance between them is \( 2d \). The midpoint \( O \) is where the charge \( +q \) is placed.

### Step 2: Calculate the Distance from \( O \) to Each Charge
Since \( O \) is the midpoint, the distance from \( O \) to each charge \( -q \) is:
\[ d_A = d \quad \text{(distance from O to A)} \]
\[ d_B = d \quad \text{(distance from O to B)} \]

### Step 3: Calculate the Electric Potential at Point \( O \)
The electric potential \( V \) at point \( O \) due to a point charge is given by:
\[ V = \frac{k \cdot q}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \) and \( r \) is the distance from the charge to the point where we are calculating the potential.

For the two charges \( -q \):
\[ V_O = V_A + V_B = \frac{-kq}{d} + \frac{-kq}{d} = \frac{-2kq}{d} \]

### Step 4: Calculate the Potential Energy of Charge \( +q \)
The potential energy \( U \) of a charge \( +q \) in an electric potential \( V \) is given by:
\[ U = q \cdot V \]
Substituting the expression for \( V_O \):
\[ U = +q \cdot \left(\frac{-2kq}{d}\right) = \frac{-2kq^2}{d} \]

### Step 5: Introduce a Small Displacement \( x \) from \( O \)
Now, let’s consider a small displacement \( x \) from \( O \). The new distances to the charges become:
- Distance to charge at \( A \): \( d - x \)
- Distance to charge at \( B \): \( d + x \)

### Step 6: Calculate the New Electric Potential at \( O \) with Displacement \( x \)
The new potential \( V \) at the displaced position is:
\[ V = \frac{-kq}{d - x} + \frac{-kq}{d + x} \]
Combining these gives:
\[ V = -kq \left( \frac{1}{d - x} + \frac{1}{d + x} \right) \]
Using the formula for the sum of fractions:
\[ V = -kq \left( \frac{(d + x) + (d - x)}{(d - x)(d + x)} \right) = -kq \left( \frac{2d}{d^2 - x^2} \right) \]

### Step 7: Calculate the Potential Energy with Displacement \( x \)
The potential energy \( U(x) \) when the charge is displaced by \( x \):
\[ U(x) = +q \cdot V = -\frac{2kq^2}{d^2 - x^2} \]

### Step 8: Analyze Stability
To determine the stability of the equilibrium at \( x = 0 \), we need to find the second derivative of \( U(x) \):
1. First, find the first derivative:
\[ \frac{dU}{dx} = \frac{d}{dx}\left(-\frac{2kq^2}{d^2 - x^2}\right) \]
Using the quotient rule, we find that \( \frac{dU}{dx} = 0 \) at \( x = 0 \).

2. Now, find the second derivative:
\[ \frac{d^2U}{dx^2} \text{ at } x = 0 \]
If \( \frac{d^2U}{dx^2} < 0 \), the equilibrium is unstable.

### Conclusion
After performing the calculations, we find that \( \frac{d^2U}{dx^2} < 0 \) at \( x = 0 \), indicating that the charge \( +q \) at point \( O \) is in an unstable equilibrium.