Two point charges 2q and 8q are placed at a distance r apart. Where should a third charge `-q` be placed between them so that the electrical potential energy of the system is a minimum

Options

A. at a distance of r/3 from 2q
B. at a distance of 2r/3 from 2q
C. at a distance of r/16 from 2q
D. None of the above

Answer

To solve the problem of where to place a third charge `-q` between two point charges `2q` and `8q` such that the electrical potential energy of the system is minimized, we can follow these steps:

### Step 1: Define the Positions
Let the charge `2q` be at position \(0\) and the charge `8q` be at position \(r\). We will place the charge `-q` at a distance \(x\) from the charge `2q`. Therefore, the distance from `-q` to `8q` will be \(r - x\).

### Step 2: Write the Expression for Potential Energy
The potential energy \(U\) of the system can be expressed as the sum of the potential energies due to each pair of charges:
\[ U = U_{2q, -q} + U_{8q, -q} + U_{2q, 8q} \]
Where:
- \(U_{2q, -q} = k \cdot \frac{2q \cdot (-q)}{x} = -\frac{2kq^2}{x}\)
- \(U_{8q, -q} = k \cdot \frac{8q \cdot (-q)}{r - x} = -\frac{8kq^2}{r - x}\)
- \(U_{2q, 8q} = k \cdot \frac{2q \cdot 8q}{r} = \frac{16kq^2}{r}\)

Thus, the total potential energy \(U\) becomes:
\[ U = -\frac{2kq^2}{x} - \frac{8kq^2}{r - x} + \frac{16kq^2}{r} \]

### Step 3: Differentiate the Potential Energy
To find the position \(x\) that minimizes the potential energy, we differentiate \(U\) with respect to \(x\) and set the derivative equal to zero:
\[ \frac{dU}{dx} = \frac{2kq^2}{x^2} - \frac{8kq^2}{(r - x)^2} = 0 \]

### Step 4: Solve the Equation
Rearranging gives:
\[ \frac{2}{x^2} = \frac{8}{(r - x)^2} \]
Cross-multiplying leads to:
\[ 2(r - x)^2 = 8x^2 \]
Simplifying this gives:
\[ (r - x)^2 = 4x^2 \]
Taking the square root of both sides results in two cases:
1. \(r - x = 2x\)
2. \(r - x = -2x\)

From the first case:
\[ r = 3x \implies x = \frac{r}{3} \]

From the second case:
\[ r = -x \implies x = -r \quad (\text{not a valid solution since } x \text{ must be positive}) \]

### Step 5: Conclusion
Thus, the charge `-q` should be placed at a distance of \( \frac{r}{3} \) from the charge `2q`.

### Final Answer
The third charge `-q` should be placed at a distance of \( \frac{r}{3} \) from the charge `2q`.

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