Question

A source charge generates an electric field of 4286 N/C at a distance of 2.5 m. What is the magnitude of the source charge? (Use k=)

1.2 µC
3.0 µC
1.2 C
3.0 C

Answer 1

To calculate the magnitude of the source charge \( Q \) that generates an electric field \( E \) at a distance \( r \), we can use the formula for the electric field produced by a point charge:

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

Where:
- \( E \) is the electric field (in N/C),
- \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( Q \) is the magnitude of the source charge (in C),
- \( r \) is the distance from the charge (in m).

Given:
- \( E = 4286 \, \text{N/C} \)
- \( r = 2.5 \, \text{m} \)

We can rearrange the formula to solve for \( Q \):

\[
Q = \frac{E r^2}{k}
\]

Now, substituting the given values:

1. Calculate \( r^2 \):
\[
r^2 = (2.5 \, \text{m})^2 = 6.25 \, \text{m}^2
\]

2. Substitute values into the equation for \( Q \):
\[
Q = \frac{4286 \, \text{N/C} \times 6.25 \, \text{m}^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2}
\]

Calculating the numerator:
\[
4286 \times 6.25 = 26787.5 \, \text{N m}^2/\text{C}
\]

Now divide by \( k \):
\[
Q = \frac{26787.5 \, \text{N m}^2/\text{C}}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2}
\]

3. Calculate \( Q \):
\[
Q \approx \frac{26787.5}{8.99 \times 10^9} \approx 2.976 \times 10^{-6} \, \text{C} = 2.976 \, \mu\text{C}
\]

The closest option provided is:

**3.0 µC**.

Thus, the magnitude of the source charge is approximately \( \boxed{3.0 \, \mu\text{C}} \).