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.-

Options

A. `2 m` from `q_1`
B. `1 m` from `q_1`
C. `2m` from `q_2`
D. `0.5 m` from `q_1`

Explanation

To find the point on the line between two positive point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 4 \, \mu C \) that is \( 3.0 \, m \) apart, where the net electric field is zero, we can follow these steps:

### Step 1: Understand the setup
We have two positive charges:
- Charge \( q_1 = 16 \, \mu C \) located at point A (let's say at position 0 m).
- Charge \( q_2 = 4 \, \mu C \) located at point B (at position 3 m).

We need to find a point \( P \) on the line between these charges where the electric fields due to both charges cancel each other out.

### Step 2: Define the position of point P
Let the distance from charge \( q_1 \) to point \( P \) be \( r \). Therefore, the distance from charge \( q_2 \) to point \( P \) will be \( 3 - r \).

### Step 3: Write the expressions for electric fields
The electric field \( E_1 \) due to charge \( q_1 \) at point \( P \) is given by:
\[ E_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1}{r^2} \]
The electric field \( E_2 \) due to charge \( q_2 \) at point \( P \) is given by:
\[ E_2 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_2}{(3 - r)^2} \]

### 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:
\[ \frac{q_1}{r^2} = \frac{q_2}{(3 - r)^2} \]

### Step 5: Substitute the values of charges
Substituting \( q_1 = 16 \times 10^{-6} \, C \) and \( q_2 = 4 \times 10^{-6} \, C \):
\[ \frac{16 \times 10^{-6}}{r^2} = \frac{4 \times 10^{-6}}{(3 - r)^2} \]

### Step 6: Simplify the equation
We can cancel \( 10^{-6} \) from both sides:
\[ \frac{16}{r^2} = \frac{4}{(3 - r)^2} \]
Cross-multiplying gives us:
\[ 16(3 - r)^2 = 4r^2 \]

### Step 7: Expand and rearrange the equation
Expanding the left side:
\[ 16(9 - 6r + r^2) = 4r^2 \]
This simplifies to:
\[ 144 - 96r + 16r^2 = 4r^2 \]
Rearranging gives:
\[ 12r^2 - 96r + 144 = 0 \]

### Step 8: Solve the quadratic equation
Dividing the entire equation by 12:
\[ r^2 - 8r + 12 = 0 \]
Factoring the quadratic:
\[ (r - 6)(r - 2) = 0 \]
Thus, \( r = 6 \) or \( r = 2 \).

### Step 9: Determine the valid solution
Since \( r \) must be between the two charges (0 to 3 m), we take \( r = 2 \, m \).

### Step 10: Find the position of point P
The position of point \( P \) is \( 2 \, m \) from charge \( q_1 \) and \( 1 \, m \) from charge \( q_2 \).

### Final Answer
The point on the line between the charges where the net electric field is zero is \( 2 \, m \) from \( q_1 \) (16 µC) and \( 1 \, m \) from \( q_2 \) (4 µC).

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