Certainly! Let's solve the problem step by step to find the magnitude of the electric field.
Step 1: Understanding the Problem
We have a charge of [tex]\(-2 \mu C\)[/tex] (microCoulombs) and it is placed at a distance of 0.5 meters. We need to calculate the magnitude of the electric field created by this charge at that distance.
Step 2: Converting Units
First, we need to convert the charge [tex]\(-2 \mu C\)[/tex] to Coulombs because 1 microCoulomb ([tex]\(\mu C\)[/tex]) is [tex]\(1 \times 10^{-6}\)[/tex] Coulombs (C).
[tex]\[
-2 \mu C = -2 \times 10^{-6} \text{ C}
\][/tex]
Step 3: Using Coulomb's Law
Coulomb's law states that the magnitude of the electric field (E) created by a point charge (q) at a distance (r) is given by the formula:
[tex]\[
E = \frac{k \cdot |q|}{r^2}
\][/tex]
where:
- [tex]\(k\)[/tex] is Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)[/tex])
- [tex]\(|q|\)[/tex] is the absolute value of the charge
- [tex]\(r\)[/tex] is the distance from the charge
Step 4: Plugging in the Values
Given:
- [tex]\(|q| = 2 \times 10^{-6} \text{ C}\)[/tex] (the absolute value of [tex]\(-2 \mu C\)[/tex])
- [tex]\(r = 0.5 \text{ m}\)[/tex]
- [tex]\(k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)[/tex]
Substitute these values into the formula:
[tex]\[
E = \frac{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \cdot 2 \times 10^{-6} \text{ C}}{(0.5 \text{ m})^2}
\][/tex]
Step 5: Simplifying the Calculation
Calculate the denominator first:
[tex]\[
(0.5 \text{ m})^2 = 0.25 \text{ m}^2
\][/tex]
Then the entire expression:
[tex]\[
E = \frac{8.99 \times 10^9 \cdot 2 \times 10^{-6}}{0.25}
\][/tex]
[tex]\[
E = \frac{17.98 \times 10^3}{0.25}
\][/tex]
Finally, divide:
[tex]\[
E = 71920 \, \text{N/C}
\][/tex]
Therefore, the magnitude of the electric field is [tex]\(71,920 \, \text{N/C}\)[/tex].
