Two charge `Q_(1)=18muC` and `Q_(2)=-2 mu C` are separated by a distance R, and `Q_(1)` is on the left of `Q_(2)`. The distance of the point where the net electric field is zero is

Options

A. between `Q_(1)` and `Q_(2)`
B. left of `Q_(1)at R//2`
C. right of `Q_(2)` at R
D. right of `Q_(2)` at `R//2`

Explanation

To find the point where the net electric field is zero due to two charges \( Q_1 = 18 \, \mu C \) and \( Q_2 = -2 \, \mu C \) separated by a distance \( R \), we can follow these steps:

### Step 1: Understand the Configuration
- We have two charges, \( Q_1 \) (positive) on the left and \( Q_2 \) (negative) on the right.
- The distance between the charges is \( R \).

### Step 2: Identify Possible Locations for Zero Electric Field
- The electric field due to a positive charge points away from the charge, while the electric field due to a negative charge points towards the charge.
- The net electric field can potentially be zero either between the charges or outside the charges.

### Step 3: Analyze the Region Between the Charges
- If we consider a point \( P \) between \( Q_1 \) and \( Q_2 \), the electric fields due to both charges will point in the same direction (away from \( Q_1 \) and towards \( Q_2 \)), meaning they will add up and not cancel each other out. Therefore, the electric field cannot be zero between the charges.

### Step 4: Analyze the Region Outside the Charges
- Consider a point \( P_3 \) to the right of \( Q_2 \). Here, the electric field due to \( Q_1 \) will point to the left (away from \( Q_1 \)), and the electric field due to \( Q_2 \) will also point to the left (towards \( Q_2 \)).
- Since \( Q_1 \) is larger in magnitude than \( Q_2 \), we can set up the equation for the electric fields.

### Step 5: Set Up the Equation
- Let the distance from \( Q_2 \) to point \( P_3 \) be \( x \). The distance from \( Q_1 \) to point \( P_3 \) will then be \( R + x \).
- The electric field \( E_1 \) due to \( Q_1 \) at point \( P_3 \) is given by:
\[ E_1 = k \frac{Q_1}{(R + x)^2} \]
- The electric field \( E_2 \) due to \( Q_2 \) at point \( P_3 \) is given by:
\[ E_2 = k \frac{|Q_2|}{x^2} \]

### Step 6: Set the Electric Fields Equal
- For the net electric field to be zero:
\[ E_1 = E_2 \]
\[ k \frac{Q_1}{(R + x)^2} = k \frac{|Q_2|}{x^2} \]
- Cancel \( k \) from both sides:
\[ \frac{Q_1}{(R + x)^2} = \frac{|Q_2|}{x^2} \]

### Step 7: Substitute the Values
- Substituting \( Q_1 = 18 \times 10^{-6} \, C \) and \( |Q_2| = 2 \times 10^{-6} \, C \):
\[ \frac{18 \times 10^{-6}}{(R + x)^2} = \frac{2 \times 10^{-6}}{x^2} \]
- Cross-multiplying gives:
\[ 18 x^2 = 2 (R + x)^2 \]

### Step 8: Expand and Rearrange
- Expanding the right side:
\[ 18 x^2 = 2 (R^2 + 2Rx + x^2) \]
\[ 18 x^2 = 2R^2 + 4Rx + 2x^2 \]
- Rearranging gives:
\[ 16 x^2 - 4Rx - 2R^2 = 0 \]

### Step 9: Solve the Quadratic Equation
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 16 \), \( b = -4R \), and \( c = -2R^2 \).
- Calculate the discriminant:
\[ b^2 - 4ac = (-4R)^2 - 4 \cdot 16 \cdot (-2R^2) = 16R^2 + 128R^2 = 144R^2 \]
- Thus:
\[ x = \frac{4R \pm 12R}{32} \]
- This gives two solutions:
\[ x = \frac{16R}{32} = \frac{R}{2} \quad \text{(valid solution)} \]
\[ x = \frac{-8R}{32} \quad \text{(not valid)} \]

### Conclusion
- The point where the net electric field is zero is at a distance \( \frac{R}{2} \) from \( Q_2 \) towards the right.