A point charge q produces an electric field of magnitude `2.0 NC^(-1)` at a point distant 50 cm from it. From the value of q.

Answer

To find the value of the point charge \( q \) that produces an electric field of magnitude \( 2.0 \, \text{N/C} \) at a distance of \( 50 \, \text{cm} \) (or \( 0.5 \, \text{m} \)), we can use the formula for the electric field due to a point charge:

\[ E = \frac{k \cdot |q|}{r^2} \]

Where:
- \( E \) is the electric field (in N/C),
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( |q| \) is the magnitude of the charge (in C),
- \( r \) is the distance from the charge (in m).

### Step-by-Step Solution:

1. **Convert the distance from cm to m**:
\[ r = 50 \, \text{cm} = \frac{50}{100} \, \text{m} = 0.5 \, \text{m} \]

2. **Use the formula for electric field**:
\[ E = \frac{k \cdot |q|}{r^2} \]

3. **Rearrange the formula to solve for \( |q| \)**:
\[ |q| = \frac{E \cdot r^2}{k} \]

4. **Substitute the known values into the equation**:
- \( E = 2.0 \, \text{N/C} \)
- \( r = 0.5 \, \text{m} \)
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

\[ |q| = \frac{2.0 \, \text{N/C} \cdot (0.5 \, \text{m})^2}{9 \times 10^9 \, \text{N m}^2/\text{C}^2} \]

5. **Calculate \( r^2 \)**:
\[ (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2 \]

6. **Substitute \( r^2 \) back into the equation**:
\[ |q| = \frac{2.0 \, \text{N/C} \cdot 0.25 \, \text{m}^2}{9 \times 10^9 \, \text{N m}^2/\text{C}^2} \]

7. **Calculate the numerator**:
\[ 2.0 \cdot 0.25 = 0.5 \]

8. **Now substitute this value**:
\[ |q| = \frac{0.5}{9 \times 10^9} \]

9. **Perform the division**:
\[ |q| = \frac{0.5}{9} \times 10^{-9} = 0.055555... \times 10^{-9} \, \text{C} \]

10. **Convert to scientific notation**:
\[ |q| \approx 5.56 \times 10^{-11} \, \text{C} \]

### Final Answer:
The value of the charge \( q \) is approximately \( 5.56 \times 10^{-11} \, \text{C} \).