Two charge 9e 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 9e charge
B. `((r )/(1+sqrt(1//3)))` from 9e charge
C. `((r )/(1-sqrt(3)))` from 3e charge
D. `((r )/(1+sqrt(1//3)))` from 3e charge

Answer

To find the distance from the charge \(9e\) where the electric field intensity will be zero, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- We have two charges: \(Q_1 = 9e\) and \(Q_2 = 3e\).
- The distance between the two charges is \(r\).

2. **Set Up the Problem**:
- We want to find a point where the electric field due to both charges cancels out. This point can either be between the charges or outside them.

3. **Consider the Electric Field Directions**:
- The electric field due to a positive charge points away from the charge.
- Therefore, the electric field due to \(Q_1\) (9e) will point away from it, and the electric field due to \(Q_2\) (3e) will also point away from it.

4. **Assume a Point Between the Charges**:
- Let \(x\) be the distance from the charge \(Q_1\) (9e) to the point where the electric field is zero.
- The distance from \(Q_2\) (3e) to this point will then be \(r - x\).

5. **Write the Expression for Electric Fields**:
- The electric field \(E_1\) due to charge \(Q_1\) at distance \(x\) is given by:
\[ E_1 = \frac{k \cdot 9e}{x^2} \]
- The electric field \(E_2\) due to charge \(Q_2\) at distance \(r - x\) is given by:
\[ E_2 = \frac{k \cdot 3e}{(r - x)^2} \]

6. **Set the Electric Fields Equal**:
- For the electric field to be zero, the magnitudes of the electric fields must be equal:
\[ \frac{9e}{x^2} = \frac{3e}{(r - x)^2} \]

7. **Cancel Common Terms**:
- We can cancel \(e\) and \(k\) from both sides:
\[ \frac{9}{x^2} = \frac{3}{(r - x)^2} \]

8. **Cross Multiply**:
- Cross multiplying gives:
\[ 9(r - x)^2 = 3x^2 \]

9. **Expand and Rearrange**:
- Expanding the left side:
\[ 9(r^2 - 2rx + x^2) = 3x^2 \]
- Rearranging gives:
\[ 9r^2 - 18rx + 9x^2 - 3x^2 = 0 \]
- This simplifies to:
\[ 9r^2 - 18rx + 6x^2 = 0 \]

10. **Use the Quadratic Formula**:
- This is a quadratic equation in \(x\):
\[ 6x^2 - 18rx + 9r^2 = 0 \]
- Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{18r \pm \sqrt{(-18r)^2 - 4 \cdot 6 \cdot 9r^2}}{2 \cdot 6} \]
\[ x = \frac{18r \pm \sqrt{324r^2 - 216r^2}}{12} \]
\[ x = \frac{18r \pm \sqrt{108r^2}}{12} \]
\[ x = \frac{18r \pm 6\sqrt{3}r}{12} \]
\[ x = \frac{3r(3 \pm \sqrt{3})}{6} = \frac{r(3 \pm \sqrt{3})}{2} \]

11. **Select the Positive Solution**:
- Since \(x\) must be positive, we take the positive root:
\[ x = \frac{r(3 - \sqrt{3})}{2} \]

### Final Answer:
The distance from the charge \(9e\) where the electric field intensity will be zero is:
\[ x = \frac{r(3 - \sqrt{3})}{2} \]