Two charges 2 nano coulombs and -6 nano coulombs are separated by 16 cm in air. The resultant electric intensity at the zero potential point which lies in between them and on the line joining them is

Options

A. `15000 NC^(-1)`
B. `7500 NC^(-1)`
C. `450 NC^(-1)`
D. `1.5 NC^(-1)`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Problem
We have two charges:
- \( q_1 = 2 \, \text{nC} = 2 \times 10^{-9} \, \text{C} \)
- \( q_2 = -6 \, \text{nC} = -6 \times 10^{-9} \, \text{C} \)

These charges are separated by a distance of \( d = 16 \, \text{cm} = 0.16 \, \text{m} \). We need to find the electric field intensity at the point where the electric potential is zero, which lies between the two charges.

### Step 2: Set Up the Equation for Electric Potential
The electric potential \( V \) at a point due to a charge is given by:
\[ V = k \cdot \frac{q}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

At the zero potential point, the total potential due to both charges must be zero:
\[ V = V_1 + V_2 = 0 \]
Substituting the potentials due to both charges:
\[ k \cdot \frac{2 \times 10^{-9}}{x} + k \cdot \frac{-6 \times 10^{-9}}{0.16 - x} = 0 \]
We can factor out \( k \) and rearrange:
\[ \frac{2}{x} - \frac{6}{0.16 - x} = 0 \]

### Step 3: Solve for \( x \)
Rearranging the equation gives:
\[ \frac{2}{x} = \frac{6}{0.16 - x} \]
Cross-multiplying yields:
\[ 2(0.16 - x) = 6x \]
Expanding and rearranging:
\[ 0.32 - 2x = 6x \implies 0.32 = 8x \implies x = \frac{0.32}{8} = 0.04 \, \text{m} \]

### Step 4: Calculate the Distance from Each Charge
The distance from \( q_1 \) (2 nC) to the zero potential point is \( x = 0.04 \, \text{m} \) and from \( q_2 \) (-6 nC) is:
\[ 0.16 - x = 0.16 - 0.04 = 0.12 \, \text{m} \]

### Step 5: Calculate the Electric Field Intensity
The electric field intensity \( E \) due to a point charge is given by:
\[ E = k \cdot \frac{|q|}{r^2} \]
Calculating the electric field due to both charges at the zero potential point:

1. **For \( q_1 \)**:
\[ E_1 = k \cdot \frac{2 \times 10^{-9}}{(0.04)^2} = 9 \times 10^9 \cdot \frac{2 \times 10^{-9}}{0.0016} = 11250 \, \text{N/C} \]

2. **For \( q_2 \)**:
\[ E_2 = k \cdot \frac{6 \times 10^{-9}}{(0.12)^2} = 9 \times 10^9 \cdot \frac{6 \times 10^{-9}}{0.0144} = 37500 \, \text{N/C} \]

### Step 6: Determine the Direction of Electric Fields
- The electric field due to \( q_1 \) (positive charge) is directed away from the charge (to the right).
- The electric field due to \( q_2 \) (negative charge) is directed towards the charge (to the left).

### Step 7: Calculate the Resultant Electric Field
Since \( E_1 \) is to the right and \( E_2 \) is to the left:
\[ E_{\text{resultant}} = E_2 - E_1 = 37500 - 11250 = 26250 \, \text{N/C} \]

### Step 8: Final Result
The resultant electric intensity at the zero potential point is:
\[ E = 26250 \, \text{N/C} \]