Two charges e and 3e are placed at a distance r. The distance of the point where the electric field intensity will be zero is

Options

A. `( r )/((1+sqrt(3)))` from 3e charge
B. `( r )/((1+sqrt(3)))` from e charge
C. `( r )/((1-sqrt(3)))` from 3e charge
D. `( r )/(1+sqrt((1)/(3)))` from e charge

Answer

To find the distance from the charge \( e \) where the electric field intensity is zero, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- We have two charges: \( q_1 = e \) and \( q_2 = 3e \).
- The distance between them is \( r \).

2. **Set Up the Problem**:
- Let \( x \) be the distance from charge \( e \) where the electric field intensity is zero.
- Therefore, the distance from charge \( 3e \) will be \( r - x \).

3. **Write the Expression for Electric Fields**:
- The electric field \( E_1 \) due to charge \( e \) at distance \( x \) is given by:
\[ E_1 = \frac{k \cdot e}{x^2} \]
- The electric field \( E_2 \) due to charge \( 3e \) at distance \( r - x \) is given by:
\[ E_2 = \frac{k \cdot 3e}{(r - x)^2} \]

4. **Set the Electric Fields Equal**:
- For the electric field to be zero at that point, the magnitudes of the electric fields must be equal:
\[ E_1 = E_2 \]
- This gives us the equation:
\[ \frac{k \cdot e}{x^2} = \frac{k \cdot 3e}{(r - x)^2} \]

5. **Cancel Common Terms**:
- We can cancel \( k \) and \( e \) from both sides (assuming \( e \neq 0 \)):
\[ \frac{1}{x^2} = \frac{3}{(r - x)^2} \]

6. **Cross Multiply**:
- Cross multiplying gives:
\[ (r - x)^2 = 3x^2 \]

7. **Expand and Rearrange**:
- Expanding the left side:
\[ r^2 - 2rx + x^2 = 3x^2 \]
- Rearranging gives:
\[ r^2 - 2rx - 2x^2 = 0 \]

8. **Use the Quadratic Formula**:
- This is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \):
- Here, \( a = -2 \), \( b = -2r \), and \( c = r^2 \).
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{2r \pm \sqrt{(-2r)^2 - 4(-2)(r^2)}}{2(-2)} \]
- Simplifying gives:
\[ x = \frac{2r \pm \sqrt{4r^2 + 8r^2}}{-4} = \frac{2r \pm \sqrt{12r^2}}{-4} \]
- This simplifies to:
\[ x = \frac{2r \pm 2\sqrt{3}r}{-4} = \frac{r(1 \pm \sqrt{3})}{-2} \]

9. **Select the Positive Solution**:
- Since \( x \) must be positive, we take the negative sign:
\[ x = \frac{r(1 - \sqrt{3})}{-2} \]
- This gives us:
\[ x = \frac{r(1 + \sqrt{3})}{2} \]

10. **Final Result**:
- The distance from charge \( e \) where the electric field intensity is zero is:
\[ x = \frac{r}{\sqrt{3} + 1} \]