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

1 answer

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.