A charge produce an electric field of 1 N/C at a point distant 0.1 m from it. The magnitude of charge is

Options

A. `1.11xx10^(-12) C`
B. `9.11xx10^(-12) C`
C. `7.11xx10^(-6) C`
D. None of these

Answer

To find the magnitude of the charge that produces an electric field of 1 N/C at a distance of 0.1 m, we can use the formula for the electric field \( E \) due to a point charge \( Q \):

\[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r^2} \]

Where:
- \( E \) is the electric field (in N/C),
- \( Q \) is the charge (in coulombs),
- \( r \) is the distance from the charge (in meters),
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

Given:
- \( E = 1 \, \text{N/C} \)
- \( r = 0.1 \, \text{m} \)

### Step 1: Rearranging the formula
We can rearrange the formula to solve for \( Q \):

\[ Q = E \cdot 4 \pi \epsilon_0 \cdot r^2 \]

### Step 2: Substituting the known values
Now, substitute the known values into the equation:

\[ Q = 1 \cdot 4 \pi (8.85 \times 10^{-12}) \cdot (0.1)^2 \]

### Step 3: Calculating \( r^2 \)
Calculate \( r^2 \):

\[ r^2 = (0.1)^2 = 0.01 \, \text{m}^2 \]

### Step 4: Calculating \( 4 \pi \epsilon_0 \)
Now calculate \( 4 \pi \epsilon_0 \):

\[ 4 \pi \epsilon_0 \approx 4 \cdot 3.14 \cdot (8.85 \times 10^{-12}) \approx 1.112 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \]

### Step 5: Final calculation for \( Q \)
Now substitute this back into the equation for \( Q \):

\[ Q = 1 \cdot (1.112 \times 10^{-10}) \cdot 0.01 \]

\[ Q = 1.112 \times 10^{-12} \, \text{C} \]

### Conclusion
The magnitude of the charge \( Q \) is approximately:

\[ Q \approx 1.11 \times 10^{-12} \, \text{C} \]