A point charge of `10 mu C` is placed at the origin.At what location on the `X` -axis should a point charge of `40mu C` be placed so that the net electric field is zero at `x=2cm` on the `X` -axis?

Options

A. `x=-4cm`
B. `x=4cm`
C. `x=8cm`
D. `x=6cm`

Answer

To solve the problem, we need to find the position on the x-axis where a charge of \(40 \mu C\) should be placed such that the net electric field at \(x = 2 \, cm\) is zero.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a point charge \(q_1 = 10 \mu C\) located at the origin \((0, 0)\).
- We want to find the position \(x\) for another charge \(q_2 = 40 \mu C\) on the x-axis such that the electric field at the point \(x = 2 \, cm\) is zero.

2. **Electric Field Due to a Point Charge**:
- The electric field \(E\) due to a point charge is given by the formula:
\[ E = k \frac{|q|}{r^2} \]
where \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge to the point where the field is being calculated.

3. **Calculating the Electric Field at \(x = 2 \, cm\)**:
- The distance from the charge \(q_1\) at the origin to the point \(x = 2 \, cm\) is:
\[ r_1 = 2 \, cm = 0.02 \, m \]
- The electric field at \(x = 2 \, cm\) due to \(q_1\) is:
\[ E_1 = k \frac{10 \times 10^{-6}}{(0.02)^2} \]

4. **Position of Charge \(q_2\)**:
- Let the position of the charge \(q_2\) be at \(x = a \, cm\).
- The distance from \(q_2\) to the point \(x = 2 \, cm\) is:
\[ r_2 = |a - 2| \, cm \]
- The electric field at \(x = 2 \, cm\) due to \(q_2\) is:
\[ E_2 = k \frac{40 \times 10^{-6}}{(a - 0.02)^2} \]

5. **Setting Up the Equation for Zero Electric Field**:
- For the net electric field to be zero at \(x = 2 \, cm\), the magnitudes of the electric fields due to both charges must be equal:
\[ E_1 = E_2 \]
- This gives us:
\[ k \frac{10 \times 10^{-6}}{(0.02)^2} = k \frac{40 \times 10^{-6}}{(a - 0.02)^2} \]
- We can cancel \(k\) and rearrange:
\[ \frac{10}{(0.02)^2} = \frac{40}{(a - 0.02)^2} \]

6. **Cross-Multiplying and Simplifying**:
- Cross-multiplying gives:
\[ 10(a - 0.02)^2 = 40(0.02)^2 \]
- Simplifying further:
\[ (a - 0.02)^2 = 4(0.02)^2 \]
- Taking the square root:
\[ a - 0.02 = \pm 0.08 \]

7. **Finding Possible Values for \(a\)**:
- This gives two possible solutions:
\[ a = 0.02 + 0.08 = 0.10 \, m \quad \text{(or 10 cm)} \]
\[ a = 0.02 - 0.08 = -0.06 \, m \quad \text{(not valid since it is negative)} \]

8. **Final Result**:
- The position of the charge \(q_2\) should be at \(x = 10 \, cm\) on the x-axis.