The two point charges `4muC` and `1muC` are separated by a distance of 2 m in air. Find the point on the line joining the charges at which net electric field of the system is zero.

Options

A. `4/3 m` from `1muC`
B. `2/3 m` from `1muC`
C. `4/3 m` from `4muC`
D. `2/3 m` from `4muC`

Explanation

To find the point on the line joining two point charges where the net electric field is zero, we can follow these steps:

### Step 1: Understand the Configuration
We have two point charges:
- Charge \( q_1 = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- Charge \( q_2 = 1 \, \mu C = 1 \times 10^{-6} \, C \)

The distance between the charges is \( d = 2 \, m \).

### Step 2: Identify the Points of Interest
The point where the electric field is zero must lie on the line joining the two charges. There are two possible regions to consider:
1. To the left of \( q_1 \)
2. Between \( q_1 \) and \( q_2 \)
3. To the right of \( q_2 \)

### Step 3: Analyze the Regions
1. **To the left of \( q_1 \)**: The electric field due to both charges will point towards \( q_1 \) and away from \( q_2 \). Hence, the net electric field cannot be zero here.

2. **Between \( q_1 \) and \( q_2 \)**: The electric field due to \( q_1 \) will point away from it, while the electric field due to \( q_2 \) will point towards it. This region is a candidate for the electric field to be zero.

3. **To the right of \( q_2 \)**: The electric field due to both charges will point away from both charges. Hence, the net electric field cannot be zero here.

### Step 4: Set Up the Equation
Let the point where the electric field is zero be at a distance \( x \) from \( q_1 \) (and thus \( 2 - x \) from \( q_2 \)).

The electric field \( E_1 \) due to \( q_1 \) at this point is given by:
\[ E_1 = \frac{k \cdot |q_1|}{x^2} \]
The electric field \( E_2 \) due to \( q_2 \) at this point is given by:
\[ E_2 = \frac{k \cdot |q_2|}{(2 - x)^2} \]

For the net electric field to be zero:
\[ E_1 = E_2 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \frac{k \cdot 4 \times 10^{-6}}{x^2} = \frac{k \cdot 1 \times 10^{-6}}{(2 - x)^2} \]

### Step 5: Simplify the Equation
We can cancel \( k \) from both sides:
\[ \frac{4 \times 10^{-6}}{x^2} = \frac{1 \times 10^{-6}}{(2 - x)^2} \]
Cross-multiplying gives:
\[ 4 \times (2 - x)^2 = 1 \times x^2 \]
Expanding and simplifying:
\[ 4(4 - 4x + x^2) = x^2 \]
\[ 16 - 16x + 4x^2 = x^2 \]
\[ 3x^2 - 16x + 16 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -16, c = 16 \):
\[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 16}}{2 \cdot 3} \]
\[ x = \frac{16 \pm \sqrt{256 - 192}}{6} \]
\[ x = \frac{16 \pm \sqrt{64}}{6} \]
\[ x = \frac{16 \pm 8}{6} \]
Calculating the two possible values:
1. \( x = \frac{24}{6} = 4 \, m \) (not valid, as it is outside the range)
2. \( x = \frac{8}{6} = \frac{4}{3} \, m \)

### Step 7: Conclusion
The point at which the net electric field is zero is at a distance of \( \frac{4}{3} \, m \) from the charge \( q_1 \) towards \( q_2 \).