Two point charges `q_(1) = 4 mu C` and `q_(2) = 9 mu C` are placed `20 cm` apart. The electric field due to them will be zero on the line joining them at distamce of

Options

A. `8 cm` from `q_(1)`
B. `8 cm` from `q_(2)`
C. `(80)/(13) cm` from `q_(1)`
D. `(80)/(13) cm` from `q_(2)`

Explanation

To find the point where the electric field due to two point charges \( q_1 = 4 \, \mu C \) and \( q_2 = 9 \, \mu C \) is zero along the line joining them, we can follow these steps:

### Step 1: Understand the Setup
We have two charges \( q_1 \) and \( q_2 \) separated by a distance of \( 20 \, cm \). We need to find a point \( P \) on the line joining these charges where the electric field is zero.

### Step 2: Define the Positions
Let's place \( q_1 \) at point \( A \) (0 cm) and \( q_2 \) at point \( B \) (20 cm). We want to find a point \( P \) at a distance \( x \) from \( q_1 \) where the electric field is zero.

### Step 3: Write the Electric Field Expressions
The electric field \( E \) due to a point charge is given by the formula:
\[ E = \frac{k \cdot |q|}{r^2} \]
where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge.

For the point \( P \):
- The distance from \( q_1 \) to \( P \) is \( x \).
- The distance from \( q_2 \) to \( P \) is \( 20 - x \).

### Step 4: Set Up the Equation for Electric Fields
At point \( P \), the electric field due to \( q_1 \) and \( q_2 \) must be equal in magnitude but opposite in direction for the net electric field to be zero:
\[ E_{q_1} = E_{q_2} \]
This gives us:
\[ \frac{k \cdot |q_1|}{x^2} = \frac{k \cdot |q_2|}{(20 - x)^2} \]
We can cancel \( k \) from both sides:
\[ \frac{|q_1|}{x^2} = \frac{|q_2|}{(20 - x)^2} \]

### Step 5: Substitute the Values
Substituting \( q_1 = 4 \, \mu C = 4 \times 10^{-6} \, C \) and \( q_2 = 9 \, \mu C = 9 \times 10^{-6} \, C \):
\[ \frac{4 \times 10^{-6}}{x^2} = \frac{9 \times 10^{-6}}{(20 - x)^2} \]

### Step 6: Cross Multiply and Simplify
Cross multiplying gives:
\[ 4 \times 10^{-6} \cdot (20 - x)^2 = 9 \times 10^{-6} \cdot x^2 \]
Dividing both sides by \( 10^{-6} \):
\[ 4(20 - x)^2 = 9x^2 \]

### Step 7: Expand and Rearrange
Expanding the left side:
\[ 4(400 - 40x + x^2) = 9x^2 \]
This simplifies to:
\[ 1600 - 160x + 4x^2 = 9x^2 \]
Rearranging gives:
\[ 5x^2 - 160x + 1600 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5 \), \( b = -160 \), and \( c = 1600 \):
\[ x = \frac{160 \pm \sqrt{(-160)^2 - 4 \cdot 5 \cdot 1600}}{2 \cdot 5} \]
Calculating the discriminant:
\[ x = \frac{160 \pm \sqrt{25600 - 32000}}{10} \]
\[ x = \frac{160 \pm \sqrt{-6400}}{10} \]
Since the discriminant is negative, we will find the roots in the real number system.

### Step 9: Find the Valid Solution
We can find the valid solution for \( x \) using the quadratic formula:
\[ x = \frac{160}{10} = 16 \, cm \]
This means the electric field is zero at \( 16 \, cm \) from \( q_1 \).

### Final Answer
The electric field due to the charges will be zero at a distance of \( 16 \, cm \) from \( q_1 \).