Two positive point charges `q_1=16muC` and `q_2=4muC` are separated in vacuum by a distance of `3.0m`. Find the point on the line between the charges where the net electric field is zero.-

Explanation

To find the point on the line between two positive point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 4 \, \mu C \), separated by a distance of \( 3.0 \, m \), where the net electric field is zero, follow these steps:

### Step 1: Understand the Configuration
We have two charges:
- \( q_1 = 16 \, \mu C \) located at point A.
- \( q_2 = 4 \, \mu C \) located at point B.
The distance between the charges is \( 3.0 \, m \).

### Step 2: Define the Point of Interest
Let’s denote the point where the electric field is zero as point P. Assume that point P is located at a distance \( x \) from charge \( q_2 \) (4 µC) and consequently at a distance \( 3 - x \) from charge \( q_1 \) (16 µC).

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

1. The electric field at point P due to \( q_1 \) (16 µC) is:
\[ E_1 = \frac{k \cdot 16 \times 10^{-6}}{(3 - x)^2} \]
2. The electric field at point P due to \( q_2 \) (4 µC) is:
\[ E_2 = \frac{k \cdot 4 \times 10^{-6}}{x^2} \]

### Step 4: Set the Electric Fields Equal
Since we want the net electric field at point P to be zero, we set the magnitudes of the electric fields equal to each other:
\[ E_1 = E_2 \]
This gives us:
\[ \frac{k \cdot 16 \times 10^{-6}}{(3 - x)^2} = \frac{k \cdot 4 \times 10^{-6}}{x^2} \]
We can cancel \( k \) from both sides:
\[ \frac{16 \times 10^{-6}}{(3 - x)^2} = \frac{4 \times 10^{-6}}{x^2} \]

### Step 5: Simplify the Equation
Now, we can simplify the equation:
\[ \frac{16}{(3 - x)^2} = \frac{4}{x^2} \]
Cross-multiplying gives:
\[ 16x^2 = 4(3 - x)^2 \]

### Step 6: Expand and Rearrange
Expanding the right side:
\[ 16x^2 = 4(9 - 6x + x^2) \]
\[ 16x^2 = 36 - 24x + 4x^2 \]
Now, rearranging the equation:
\[ 16x^2 - 4x^2 + 24x - 36 = 0 \]
\[ 12x^2 + 24x - 36 = 0 \]

### Step 7: Simplify the Quadratic Equation
Dividing the entire equation by 12:
\[ x^2 + 2x - 3 = 0 \]

### Step 8: Factor the Quadratic Equation
Factoring gives:
\[ (x + 3)(x - 1) = 0 \]
Thus, \( x = -3 \) or \( x = 1 \). Since \( x \) must be positive and within the range of 0 to 3 meters, we take:
\[ x = 1 \, m \]

### Step 9: Find the Distances from Each Charge
The distance from \( q_2 \) (4 µC) to point P is \( 1 \, m \), and the distance from \( q_1 \) (16 µC) to point P is:
\[ 3 - x = 3 - 1 = 2 \, m \]

### Final Answer
The point where the net electric field is zero is at a distance of \( 1 \, m \) from charge \( q_2 \) and \( 2 \, m \) from charge \( q_1 \).

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