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

1 answer

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

  1. 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 \( \boxed{3.0 , \mu\text{C}} \).