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 `25mu C`
B. At a distance of `5 cm` from `36mu C`
C. At a distance of `10 cm` from `25mu C`
D. At a distance of `11 cm` from `36mu C`

Explanation

To find the point on the line joining two charges \( 25 \, \mu C \) and \( 36 \, \mu C \) where the electric field intensity is zero, we can follow these steps:

### Step 1: Understand the Configuration
We have two positive charges:
- Charge \( Q_1 = 25 \, \mu C \)
- Charge \( Q_2 = 36 \, \mu C \)
- Distance between the charges \( d = 11 \, cm = 0.11 \, m \)

### Step 2: Identify the Regions
The electric field can be zero at three possible locations:
1. To the left of \( Q_1 \)
2. Between \( Q_1 \) and \( Q_2 \)
3. To the right of \( Q_2 \)

Since both charges are positive, the electric field due to each charge will point away from the charges. Therefore, the point where the electric field is zero must be between the two charges.

### Step 3: Set Up the Equation
Let \( x \) be the distance from \( Q_1 \) to the point where the electric field is zero. The distance from \( Q_2 \) to this point will then be \( 0.11 - x \).

The electric field \( E \) due to a point charge is given by:
\[ E = k \frac{Q}{r^2} \]
where \( k \) is Coulomb's constant.

At the point where the electric field is zero, the magnitudes of the electric fields due to both charges must be equal:
\[ E_1 = E_2 \]
Thus,
\[ k \frac{25 \times 10^{-6}}{x^2} = k \frac{36 \times 10^{-6}}{(0.11 - x)^2} \]
We can cancel \( k \) from both sides:
\[ \frac{25 \times 10^{-6}}{x^2} = \frac{36 \times 10^{-6}}{(0.11 - x)^2} \]

### Step 4: Simplify the Equation
Removing \( 10^{-6} \) from both sides gives:
\[ \frac{25}{x^2} = \frac{36}{(0.11 - x)^2} \]

Cross-multiplying results in:
\[ 25(0.11 - x)^2 = 36x^2 \]

### Step 5: Expand and Rearrange
Expanding the left side:
\[ 25(0.0121 - 0.22x + x^2) = 36x^2 \]
This simplifies to:
\[ 0.3025 - 5.5x + 25x^2 = 36x^2 \]
Rearranging gives:
\[ 25x^2 - 36x^2 - 5.5x + 0.3025 = 0 \]
\[ -11x^2 - 5.5x + 0.3025 = 0 \]
Multiplying through by -1:
\[ 11x^2 + 5.5x - 0.3025 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 11, b = 5.5, c = -0.3025 \).

Calculating the discriminant:
\[ b^2 - 4ac = (5.5)^2 - 4 \cdot 11 \cdot (-0.3025) = 30.25 + 13.33 = 43.58 \]
Calculating \( x \):
\[ x = \frac{-5.5 \pm \sqrt{43.58}}{2 \cdot 11} \]
\[ x = \frac{-5.5 \pm 6.6}{22} \]

Calculating the two possible values:
1. \( x = \frac{1.1}{22} \approx 0.05 \, m \) (valid)
2. \( x = \frac{-12.1}{22} \) (not valid as distance cannot be negative)

### Step 7: Conclusion
The point where the electric field intensity is zero is approximately \( 0.05 \, m \) or \( 5 \, cm \) from the charge \( Q_1 \).