The distance between the two charges `25 mu C` and `36 muC` is `11 cm`. At what point on the line joining the two, the intensity will be zero

Options

A. At a distance of 5 cm from `25 muC`
B. At a distance of 5 cm from `36 muC`
C. At a distance of 4 cm from `25 muC`
D. At a distance of 4 cm from `36 muC`

Answer

To solve the problem of finding the point on the line joining two charges 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:
- \( q_1 = 25 \, \mu C = 25 \times 10^{-6} \, C \)
- \( q_2 = 36 \, \mu C = 36 \times 10^{-6} \, C \)
- The distance between the two charges is \( d = 11 \, cm = 0.11 \, m \).

2. **Set Up the Problem**:
- Let \( P \) be the point where the electric field intensity is zero.
- Let \( x \) be the distance from \( q_1 \) (25 µC) to point \( P \).
- Therefore, the distance from \( q_2 \) (36 µC) to point \( P \) will be \( 11 - x \).

3. **Write the Electric Field Intensity Equations**:
- The electric field intensity due to \( q_1 \) at point \( P \) is given by:
\[ E_1 = \frac{k \cdot q_1}{x^2} \]
- The electric field intensity due to \( q_2 \) at point \( P \) is given by:
\[ E_2 = \frac{k \cdot q_2}{(11 - x)^2} \]

4. **Set the Electric Fields Equal**:
- For the net electric field at point \( P \) to be zero, we set \( E_1 = E_2 \):
\[ \frac{k \cdot q_1}{x^2} = \frac{k \cdot q_2}{(11 - x)^2} \]
- The constant \( k \) cancels out from both sides.

5. **Substitute the Values**:
- Substituting \( q_1 \) and \( q_2 \):
\[ \frac{25 \times 10^{-6}}{x^2} = \frac{36 \times 10^{-6}}{(11 - x)^2} \]

6. **Cross Multiply**:
- Cross multiplying gives:
\[ 25 \times 10^{-6} \cdot (11 - x)^2 = 36 \times 10^{-6} \cdot x^2 \]

7. **Simplify the Equation**:
- Dividing both sides by \( 10^{-6} \):
\[ 25(11 - x)^2 = 36x^2 \]

8. **Expand and Rearrange**:
- Expanding the left side:
\[ 25(121 - 22x + x^2) = 36x^2 \]
- This simplifies to:
\[ 3025 - 550x + 25x^2 = 36x^2 \]
- Rearranging gives:
\[ 0 = 11x^2 + 550x - 3025 \]

9. **Solve the Quadratic Equation**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 11, b = 550, c = -3025 \).
- Calculate the discriminant:
\[ b^2 - 4ac = 550^2 - 4 \cdot 11 \cdot (-3025) \]
- Solve for \( x \).

10. **Find the Valid Solution**:
- After calculating, we find \( x \) and determine if it's a valid solution (it should be between 0 and 11 cm).

### Final Answer:
The point where the electric field intensity is zero is \( 5 \, cm \) from the \( 25 \, \mu C \) charge.