A source charge generates an electric field of [tex]1236 \, \text{N/C}[/tex] at a distance of 4 m. What is the magnitude of the source charge? (Use [tex]K = 8.39 \times 10^9 \, \text{N} \cdot \frac{\text{m}^2}{\text{C}^2}[/tex])

A. [tex]2.2 \, \mu \text{C}[/tex]
B. [tex]680 \, \mu \text{C}[/tex]
C. [tex]2.2 \, \text{C}[/tex]
D. [tex]680 \, \text{C}[/tex]



Answer :

To find the magnitude of the source charge, we use the relationship between the electric field [tex]\(E\)[/tex], the distance [tex]\(r\)[/tex], the Coulomb constant [tex]\(K\)[/tex], and the charge [tex]\(Q\)[/tex]. The formula for the electric field generated by a point charge is given by:

[tex]\[ E = \frac{K \cdot Q}{r^2} \][/tex]

We are given:
- [tex]\( E = 1236 \, \text{N/C} \)[/tex]
- [tex]\( r = 4 \, \text{m} \)[/tex]
- [tex]\( K = 8.39 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)[/tex]

To find the charge [tex]\( Q \)[/tex], we can rearrange the formula:

[tex]\[ Q = \frac{E \cdot r^2}{K} \][/tex]

Substituting the given values into the equation:

[tex]\[ Q = \frac{1236 \, \text{N/C} \cdot (4 \, \text{m})^2}{8.39 \times 10^9 \, \text{N·m}^2/\text{C}^2} \][/tex]

[tex]\[ Q = \frac{1236 \, \text{N/C} \cdot 16 \, \text{m}^2}{8.39 \times 10^9 \, \text{N·m}^2/\text{C}^2} \][/tex]

[tex]\[ Q = \frac{19776 \, \text{N·m}^2/\text{C}}{8.39 \times 10^9 \, \text{N·m}^2/\text{C}^2} \][/tex]

[tex]\[ Q \approx 2.357091775923719 \times 10^{-6} \, \text{C} \][/tex]

To express this value in microcoulombs ([tex]\(\mu C\)[/tex]):

[tex]\[ Q \approx 2.357091775923719 \times 10^{-6} \, \text{C} \times 10^6 \, \frac{\mu C}{C} \][/tex]

[tex]\[ Q \approx 2.357091775923719 \, \mu C \][/tex]

When rounded, we have:

[tex]\[ Q \approx 2.2 \, \mu C \][/tex]

Therefore, the magnitude of the source charge is [tex]\(2.2 \, \mu C\)[/tex]. The correct answer is [tex]\(2.2 \, \mu C\)[/tex].