A charge of `+2muC` is placed at `x=0` and a charge of `-32muc` at `x=60` .A third charge `-Q` be placed on the `x` axis such that it experiences no force .The distance of the point from `+2muC` is ( in cm)

Options

A. `-20`
B. `20`
C. `15`
D. `10`

Explanation

To solve the problem, we need to find the position along the x-axis where a charge of \(-Q\) can be placed such that it experiences no net force due to the other two charges, \(+2 \mu C\) at \(x=0\) and \(-32 \mu C\) at \(x=60\) cm.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two charges:
- Charge \(Q_1 = +2 \mu C\) at \(x = 0\) cm.
- Charge \(Q_2 = -32 \mu C\) at \(x = 60\) cm.
- We need to place a third charge \(Q_3 = -Q\) at a position \(x = d\) such that the net force on it is zero.

2. **Determine the Regions**:
- The charge \(Q_3\) can be placed in three possible regions:
- To the left of \(Q_1\) (i.e., \(x < 0\))
- Between \(Q_1\) and \(Q_2\) (i.e., \(0 < x < 60\))
- To the right of \(Q_2\) (i.e., \(x > 60\))

Since \(Q_3\) is negative, it will experience a repulsive force from \(Q_1\) and an attractive force from \(Q_2\). The only region where it can experience no net force is to the left of \(Q_1\).

3. **Set Up the Electric Field Equations**:
- The electric field \(E_1\) due to \(Q_1\) at a distance \(d\) is given by:
\[ E_1 = k \frac{Q_1}{d^2} = k \frac{2 \times 10^{-6}}{d^2} \]
- The electric field \(E_2\) due to \(Q_2\) at a distance \(d + 60\) (since \(Q_2\) is at \(60\) cm) is given by:
\[ E_2 = k \frac{|Q_2|}{(d + 60)^2} = k \frac{32 \times 10^{-6}}{(d + 60)^2} \]

4. **Set the Electric Fields Equal**:
- For the charge \(Q_3\) to experience no net force, the magnitudes of the electric fields must be equal:
\[ E_1 = E_2 \]
- Therefore, we have:
\[ k \frac{2 \times 10^{-6}}{d^2} = k \frac{32 \times 10^{-6}}{(d + 60)^2} \]
- The \(k\) cancels out:
\[ \frac{2}{d^2} = \frac{32}{(d + 60)^2} \]

5. **Cross-Multiply and Simplify**:
- Cross-multiplying gives:
\[ 2(d + 60)^2 = 32d^2 \]
- Expanding the left side:
\[ 2(d^2 + 120d + 3600) = 32d^2 \]
- Simplifying:
\[ 2d^2 + 240d + 7200 = 32d^2 \]
- Rearranging:
\[ 30d^2 - 240d - 7200 = 0 \]

6. **Solve the Quadratic Equation**:
- Dividing the entire equation by 30:
\[ d^2 - 8d - 240 = 0 \]
- Using the quadratic formula \(d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ d = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-240)}}{2 \cdot 1} \]
\[ d = \frac{8 \pm \sqrt{64 + 960}}{2} \]
\[ d = \frac{8 \pm \sqrt{1024}}{2} \]
\[ d = \frac{8 \pm 32}{2} \]
- This gives two solutions:
\[ d = \frac{40}{2} = 20 \quad \text{and} \quad d = \frac{-24}{2} = -12 \]

7. **Select the Valid Solution**:
- Since we are looking for a position to the left of \(Q_1\), we take \(d = -12\) cm.
- Thus, the distance from \(+2 \mu C\) is \(12\) cm to the left.

### Final Answer:
The distance of the point from \(+2 \mu C\) is \(12\) cm (to the left).