What magnitude charge, in nano-Coulombs, creates a [tex]$36.49 \, \text{N/C}$[/tex] electric field at a point 8.38 m away?

[tex]\square[/tex]



Answer :

To determine the magnitude of the charge in nano-Coulombs that creates a [tex]\(36.49\, \text{N/C}\)[/tex] electric field at a distance of [tex]\(8.38 \, \text{m}\)[/tex], we will use the relationship between the electric field [tex]\(E\)[/tex] and the charge [tex]\(q\)[/tex], given by Coulomb's law:

[tex]\[ E = k \frac{|q|}{r^2} \][/tex]

where:
- [tex]\(E\)[/tex] is the electric field ([tex]\(\text{N/C}\)[/tex]),
- [tex]\(k\)[/tex] is Coulomb's constant (approximately [tex]\(8.99 \times 10^9 \,\text{N·m}^2/\text{C}^2\)[/tex]),
- [tex]\(r\)[/tex] is the distance from the charge ([tex]\(\text{m}\)[/tex]),
- [tex]\(q\)[/tex] is the magnitude of the charge ([tex]\(\text{C}\)[/tex]).

First, rearrange the formula to solve for the magnitude of the charge [tex]\(|q|\)[/tex]:

[tex]\[ |q| = \frac{E r^2}{k} \][/tex]

Given the values:
- [tex]\(E = 36.49\, \text{N/C}\)[/tex],
- [tex]\(r = 8.38 \, \text{m}\)[/tex],
- [tex]\(k = 8.99 \times 10^9 \,\text{N·m}^2/\text{C}^2\)[/tex],

we substitute these values into the formula:

[tex]\[ |q| = \frac{36.49 \times (8.38)^2}{8.99 \times 10^9} \][/tex]

Evaluating the numerical components step-by-step:

1. Calculate [tex]\(r^2\)[/tex]:
[tex]\[ (8.38)^2 = 70.2244 \][/tex]

2. Multiply [tex]\(E\)[/tex] by [tex]\(r^2\)[/tex]:
[tex]\[ 36.49 \times 70.2244 \approx 2563.2903 \][/tex]

3. Divide by Coulomb's constant [tex]\(k\)[/tex]:
[tex]\[ \frac{2563.2903}{8.99 \times 10^9} \approx 2.8504 \times 10^{-7} \, \text{C} \][/tex]

Thus, the magnitude of the charge in Coulombs is approximately:
[tex]\[ |q| = 2.8504 \times 10^{-7} \, \text{C} \][/tex]

To convert this charge into nano-Coulombs (nC), we use the conversion factor [tex]\(1 \, \text{C} = 10^9 \, \text{nC}\)[/tex]:

[tex]\[ \text{Charge in nC} = 2.8504 \times 10^{-7} \, \text{C} \times 10^9 \, \text{nC/C} \approx 285.0376 \, \text{nC} \][/tex]

So, the magnitude of the charge that creates a [tex]\(36.49 \, \text{N/C}\)[/tex] electric field at a point [tex]\(8.38 \, \text{m}\)[/tex] away is approximately:

[tex]\[ 285.0376 \, \text{nC} \][/tex]