The electric field due to a charge at a distance of 3 m from it is `500 NC^(-1)`. The magnitude of the charge is
`[(1)/(4 pi epsi_(0)) = 9 xx 10^(9) N - m^(2)//C^(2)]`

Options

A. `2.5 muC`
B. `2.0 muC`
C. `1.0 muC`
D. `0.5 muC`

Answer

To find the magnitude of the charge \( Q \) given the electric field \( E \) at a distance \( R \) from the charge, we can use the formula for the electric field due to a point charge:

\[ E = \frac{KQ}{R^2} \]

Where:
- \( E \) is the electric field (in N/C),
- \( K \) is Coulomb's constant, given as \( K = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( Q \) is the charge (in coulombs),
- \( R \) is the distance from the charge (in meters).

### Step-by-step Solution:

1. **Identify the given values:**
- Electric field \( E = 500 \, \text{N/C} \)
- Distance \( R = 3 \, \text{m} \)
- Coulomb's constant \( K = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

2. **Substitute the values into the electric field formula:**
\[ 500 = \frac{9 \times 10^9 \cdot Q}{3^2} \]

3. **Calculate \( R^2 \):**
\[ R^2 = 3^2 = 9 \]

4. **Rewrite the equation:**
\[ 500 = \frac{9 \times 10^9 \cdot Q}{9} \]

5. **Simplify the equation:**
\[ 500 = 10^9 \cdot Q \]

6. **Solve for \( Q \):**
\[ Q = \frac{500}{10^9} \]

7. **Convert \( Q \) into microcoulombs:**
\[ Q = 500 \times 10^{-9} \, \text{C} = 0.5 \, \mu\text{C} \]

### Final Answer:
The magnitude of the charge \( Q \) is \( 0.5 \, \mu\text{C} \).

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