Two charge `Q_(1)=18muC` and `Q_(2)=-2 mu C` are separted by a distance R, and `Q_(1)` is on the left off `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 distance from charge \( Q_1 \) where the net electric field is zero, we can follow these steps:

### Step 1: Understand the Setup
We have two charges:
- \( Q_1 = 18 \, \mu C \) (positive charge)
- \( Q_2 = -2 \, \mu C \) (negative charge)

They are separated by a distance \( R \), with \( Q_1 \) on the left and \( Q_2 \) on the right.

### Step 2: Identify the Regions for Electric Field Zero
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. Therefore, the electric field can only be zero at a point outside the two charges, specifically to the right of \( Q_2 \).

### Step 3: Set Up the Equation for Electric Fields
Let \( x \) be the distance from \( Q_2 \) to the point where the electric field is zero. The distance from \( Q_1 \) to this point will then be \( R + x \).

The electric field due to \( Q_1 \) at this point is given by:
\[ E_1 = \frac{k |Q_1|}{(R + x)^2} \]
The electric field due to \( Q_2 \) at this point is given by:
\[ E_2 = \frac{k |Q_2|}{x^2} \]

### Step 4: Set the Electric Fields Equal
For the net electric field to be zero:
\[ E_1 = E_2 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \frac{k |Q_1|}{(R + x)^2} = \frac{k |Q_2|}{x^2} \]

### Step 5: Simplify the Equation
We can cancel \( k \) from both sides:
\[ \frac{|Q_1|}{(R + x)^2} = \frac{|Q_2|}{x^2} \]
Substituting the values of \( Q_1 \) and \( Q_2 \):
\[ \frac{18 \times 10^{-6}}{(R + x)^2} = \frac{2 \times 10^{-6}}{x^2} \]

### Step 6: Cross-Multiply
Cross-multiplying gives:
\[ 18 \times 10^{-6} \cdot x^2 = 2 \times 10^{-6} \cdot (R + x)^2 \]

### Step 7: Expand and Rearrange
Expanding the right side:
\[ 18x^2 = 2(R^2 + 2Rx + x^2) \]
\[ 18x^2 = 2R^2 + 4Rx + 2x^2 \]
Rearranging gives:
\[ (18x^2 - 2x^2) - 4Rx - 2R^2 = 0 \]
\[ 16x^2 - 4Rx - 2R^2 = 0 \]

### Step 8: Solve the Quadratic Equation
This is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \):
- \( a = 16 \)
- \( b = -4R \)
- \( c = -2R^2 \)

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ x = \frac{4R \pm \sqrt{(-4R)^2 - 4 \cdot 16 \cdot (-2R^2)}}{2 \cdot 16} \]
\[ x = \frac{4R \pm \sqrt{16R^2 + 128R^2}}{32} \]
\[ x = \frac{4R \pm \sqrt{144R^2}}{32} \]
\[ x = \frac{4R \pm 12R}{32} \]
This gives two possible solutions:
1. \( x = \frac{16R}{32} = \frac{R}{2} \)
2. \( x = \frac{-8R}{32} = -\frac{R}{4} \) (not physically meaningful since distance cannot be negative)

### Final Answer
Thus, the distance from \( Q_2 \) where the electric field is zero is:
\[ x = \frac{R}{2} \]