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 point where the electric field intensity is zero due to two charges \( e \) and \( 3e \) placed at a distance \( r \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Let charge \( e \) be at point A (let's say at position 0) and charge \( 3e \) be at point B (at position \( r \)).
- We need to find the point where the electric field intensity is zero.

2. **Determine the Regions:**
- There are three regions to consider:
- Region 1: To the left of charge \( e \) (x < 0)
- Region 2: Between the two charges (0 < x < r)
- Region 3: To the right of charge \( 3e \) (x > r)

3. **Analyze Each Region:**
- **Region 1:** The electric fields due to both charges will point towards the positive charge \( 3e \) (since both charges are positive), hence they cannot cancel each other.
- **Region 2:** The electric field due to charge \( e \) will point away from it, while the electric field due to charge \( 3e \) will point towards it. This is the only region where the electric fields can potentially cancel each other.
- **Region 3:** Similar to Region 1, the electric fields will not cancel out as both will point away from \( 3e \).

4. **Set Up the Equation for Electric Fields in Region 2:**
- Let \( x \) be the distance from charge \( e \) to the point where the electric field is zero.
- The distance from charge \( 3e \) to this point will then be \( r - x \).
- The electric field due to charge \( e \) at this point is given by:
\[ E_1 = \frac{k \cdot e}{x^2} \]
- The electric field due to charge \( 3e \) at this point is given by:
\[ E_2 = \frac{k \cdot 3e}{(r - x)^2} \]

5. **Set the Electric Fields Equal to Each Other:**
- For the electric field to be zero, we set \( E_1 = E_2 \):
\[ \frac{k \cdot e}{x^2} = \frac{k \cdot 3e}{(r - x)^2} \]
- We can cancel \( k \) and \( e \) (assuming \( e \neq 0 \)):
\[ \frac{1}{x^2} = \frac{3}{(r - x)^2} \]

6. **Cross-Multiply and Rearrange:**
- Cross-multiplying gives:
\[ (r - x)^2 = 3x^2 \]
- Expanding the left side:
\[ r^2 - 2rx + x^2 = 3x^2 \]
- Rearranging gives:
\[ r^2 - 2rx - 2x^2 = 0 \]

7. **Solve the Quadratic Equation:**
- This is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \) where \( a = -2, b = -2r, c = r^2 \).
- 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:
\[ x = \frac{2r \pm \sqrt{4r^2 + 8r^2}}{-4} = \frac{2r \pm \sqrt{12r^2}}{-4} = \frac{2r \pm 2\sqrt{3}r}{-4} \]
- This simplifies to:
\[ x = \frac{r(1 \pm \sqrt{3})}{-2} \]
- We take the positive root since \( x \) must be positive:
\[ x = \frac{r(1 - \sqrt{3})}{-2} \]

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

### Final Answer:
The distance from charge \( e \) where the electric field intensity will be zero is:
\[ \frac{r}{1 + \sqrt{3}} \]