Charges `2q` and `8q` are placed at the end points `A` and `B` repsectively of a `9 cm` long straight line. A third charge `q` is placed at a point `C of AB` such that the potential energy of the system is minimum . The distance of `C` from `A` is

Options

A. `2 cm`
B. `3 cm`
C. `4cm`
D. `5 cm`

Explanation

To solve the problem, we need to find the position of charge `q` on the line segment `AB` such that the potential energy of the system is minimized. The charges `2q` and `8q` are located at points `A` and `B`, respectively, with a distance of `9 cm` between them.

### Step-by-step Solution:

1. **Define the Positions**:
- Let the position of charge `2q` at point `A` be at `x = 0 cm`.
- Let the position of charge `8q` at point `B` be at `x = 9 cm`.
- Let the position of charge `q` at point `C` be at a distance `x` from `A`, so the distance from `B` to `C` will be `9 - x`.

2. **Potential Energy Formula**:
The potential energy `U` of the system can be expressed as the sum of the potential energy contributions from each pair of charges:
\[ U = k \left( \frac{2q \cdot q}{x} + \frac{8q \cdot q}{9 - x} \right) \]
where `k` is the electrostatic constant.

3. **Differentiate the Potential Energy**:
To find the minimum potential energy, we need to take the derivative of `U` with respect to `x` and set it to zero:
\[ \frac{dU}{dx} = k \left( \frac{2q}{x^2} - \frac{8q}{(9 - x)^2} \right) = 0 \]

4. **Set the Derivative to Zero**:
Rearranging gives:
\[ \frac{2q}{x^2} = \frac{8q}{(9 - x)^2} \]
Simplifying this, we can cancel `q` (assuming `q ≠ 0`):
\[ \frac{2}{x^2} = \frac{8}{(9 - x)^2} \]

5. **Cross-Multiply**:
Cross-multiplying yields:
\[ 2(9 - x)^2 = 8x^2 \]
Expanding both sides:
\[ 2(81 - 18x + x^2) = 8x^2 \]
\[ 162 - 36x + 2x^2 = 8x^2 \]
Rearranging gives:
\[ 6x^2 - 36x + 162 = 0 \]

6. **Simplifying the Quadratic Equation**:
Dividing the entire equation by `6`:
\[ x^2 - 6x + 27 = 0 \]

7. **Using the Quadratic Formula**:
The roots can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, `a = 1`, `b = -6`, and `c = 27`:
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1} \]
\[ x = \frac{6 \pm \sqrt{36 - 108}}{2} \]
\[ x = \frac{6 \pm \sqrt{-72}}{2} \]
Since we have a negative discriminant, we need to check our previous steps for errors.

8. **Revisiting the Derivative**:
Correctly solving the equation:
\[ 9 - x = 2x \implies 9 = 3x \implies x = 3 \text{ cm} \]

### Final Answer:
The distance of point `C` from point `A` is **3 cm**.