Question
A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge? (Use k=)
2.2 µC
680 µC
2.2 C
680 C
2.2 µC
680 µC
2.2 C
680 C
Answers
GPT-4o mini
To find the magnitude of the source charge that generates a given electric field, we can use Coulomb's law. The electric field \( E \) created by a point charge \( Q \) at a distance \( r \) is given by the formula:
\[
E = \frac{k \cdot |Q|}{r^2}
\]
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 source charge (in coulombs), and
- \( r \) is the distance from the charge (in meters).
Given:
- \( E = 1236 \, \text{N/C} \)
- \( r = 4 \, \text{m} \)
We can rearrange the formula to solve for \( Q \):
\[
|Q| = \frac{E \cdot r^2}{k}
\]
Substituting the known values:
\[
|Q| = \frac{1236 \, \text{N/C} \cdot (4 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2}
\]
Calculating \( (4 \, \text{m})^2 \):
\[
(4 \, \text{m})^2 = 16 \, \text{m}^2
\]
Now substitute this back into the equation:
\[
|Q| = \frac{1236 \, \text{N/C} \cdot 16 \, \text{m}^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2}
\]
Calculating the numerator:
\[
1236 \cdot 16 = 19776
\]
Now substituting this into the equation for \( Q \):
\[
|Q| = \frac{19776}{8.99 \times 10^9}
\]
Calculating the value:
\[
|Q| \approx \frac{19776}{8.99 \times 10^9} \approx 2.20 \times 10^{-6} \, \text{C}
\]
Now, converting this into microcoulombs (\( \mu C \)):
\[
|Q| \approx 2.20 \, \mu C
\]
Therefore, the magnitude of the source charge is:
\[
\text{2.2 µC}
\]
So, the correct answer is **2.2 µC**.
\[
E = \frac{k \cdot |Q|}{r^2}
\]
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 source charge (in coulombs), and
- \( r \) is the distance from the charge (in meters).
Given:
- \( E = 1236 \, \text{N/C} \)
- \( r = 4 \, \text{m} \)
We can rearrange the formula to solve for \( Q \):
\[
|Q| = \frac{E \cdot r^2}{k}
\]
Substituting the known values:
\[
|Q| = \frac{1236 \, \text{N/C} \cdot (4 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2}
\]
Calculating \( (4 \, \text{m})^2 \):
\[
(4 \, \text{m})^2 = 16 \, \text{m}^2
\]
Now substitute this back into the equation:
\[
|Q| = \frac{1236 \, \text{N/C} \cdot 16 \, \text{m}^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2}
\]
Calculating the numerator:
\[
1236 \cdot 16 = 19776
\]
Now substituting this into the equation for \( Q \):
\[
|Q| = \frac{19776}{8.99 \times 10^9}
\]
Calculating the value:
\[
|Q| \approx \frac{19776}{8.99 \times 10^9} \approx 2.20 \times 10^{-6} \, \text{C}
\]
Now, converting this into microcoulombs (\( \mu C \)):
\[
|Q| \approx 2.20 \, \mu C
\]
Therefore, the magnitude of the source charge is:
\[
\text{2.2 µC}
\]
So, the correct answer is **2.2 µC**.