Answer :
To determine the electric field at a point 0.450 meters to the left of a [tex]\(-5.77 \times 10^{-9}\)[/tex] C charge, we utilize Coulomb's law for the electric field due to a point charge. The electric field [tex]\(E\)[/tex] created by a point charge [tex]\(q\)[/tex] at a distance [tex]\(r\)[/tex] from the charge is given by:
[tex]\[ E = k \frac{|q|}{r^2} \][/tex]
where:
- [tex]\(k\)[/tex] is Coulomb's constant, approximately [tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\)[/tex],
- [tex]\(q\)[/tex] is the charge in coulombs,
- [tex]\(r\)[/tex] is the distance from the charge to the point where the field is being calculated in meters.
Given:
[tex]\[ q = -5.77 \times 10^{-9} \, \text{C} \][/tex]
[tex]\[ r = 0.450 \, \text{m} \][/tex]
First, calculate the magnitude of the electric field:
[tex]\[ E = 8.99 \times 10^9 \frac{\text{N} \cdot \text{m}^2}{\text{C}^2} \times \frac{| - 5.77 \times 10^{-9} \text{C} |}{(0.450 \, \text{m})^2} \][/tex]
[tex]\[ |q| = 5.77 \times 10^{-9} \, \text{C} \][/tex]
[tex]\[ r^2 = (0.450)^2 = 0.2025 \, \text{m}^2 \][/tex]
Now, substituting these values into the formula:
[tex]\[ E = 8.99 \times 10^9 \frac{\text{N} \cdot \text{m}^2}{\text{C}^2} \times \frac{5.77 \times 10^{-9} \, \text{C}}{0.2025 \, \text{m}^2} \][/tex]
[tex]\[ E = 8.99 \times 10^9 \times \frac{5.77 \times 10^{-9}}{0.2025} \][/tex]
Performing the division and multiplication, we obtain the electric field magnitude:
[tex]\[ E \approx 256.1595061728395 \, \text{N/C} \][/tex]
Next, determine the direction. Since the charge is negative, the electric field direction is towards the charge. Given that the point in question is to the left of the charge, the electric field direction will also be towards the left.
Thus, the electric field at a point 0.450 meters to the left of a [tex]\(-5.77 \times 10^{-9}\)[/tex] C charge is approximately:
[tex]\[ E \approx 256.1595061728395 \, \text{N/C} \][/tex]
In conclusion, the electric field is [tex]\(256.1595061728395 \, \text{N/C}\)[/tex] directed to the left, so we include a negative sign to indicate this direction:
[tex]\[ E \approx -256.1595061728395 \, \text{N/C} \][/tex]
[tex]\[ E = k \frac{|q|}{r^2} \][/tex]
where:
- [tex]\(k\)[/tex] is Coulomb's constant, approximately [tex]\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\)[/tex],
- [tex]\(q\)[/tex] is the charge in coulombs,
- [tex]\(r\)[/tex] is the distance from the charge to the point where the field is being calculated in meters.
Given:
[tex]\[ q = -5.77 \times 10^{-9} \, \text{C} \][/tex]
[tex]\[ r = 0.450 \, \text{m} \][/tex]
First, calculate the magnitude of the electric field:
[tex]\[ E = 8.99 \times 10^9 \frac{\text{N} \cdot \text{m}^2}{\text{C}^2} \times \frac{| - 5.77 \times 10^{-9} \text{C} |}{(0.450 \, \text{m})^2} \][/tex]
[tex]\[ |q| = 5.77 \times 10^{-9} \, \text{C} \][/tex]
[tex]\[ r^2 = (0.450)^2 = 0.2025 \, \text{m}^2 \][/tex]
Now, substituting these values into the formula:
[tex]\[ E = 8.99 \times 10^9 \frac{\text{N} \cdot \text{m}^2}{\text{C}^2} \times \frac{5.77 \times 10^{-9} \, \text{C}}{0.2025 \, \text{m}^2} \][/tex]
[tex]\[ E = 8.99 \times 10^9 \times \frac{5.77 \times 10^{-9}}{0.2025} \][/tex]
Performing the division and multiplication, we obtain the electric field magnitude:
[tex]\[ E \approx 256.1595061728395 \, \text{N/C} \][/tex]
Next, determine the direction. Since the charge is negative, the electric field direction is towards the charge. Given that the point in question is to the left of the charge, the electric field direction will also be towards the left.
Thus, the electric field at a point 0.450 meters to the left of a [tex]\(-5.77 \times 10^{-9}\)[/tex] C charge is approximately:
[tex]\[ E \approx 256.1595061728395 \, \text{N/C} \][/tex]
In conclusion, the electric field is [tex]\(256.1595061728395 \, \text{N/C}\)[/tex] directed to the left, so we include a negative sign to indicate this direction:
[tex]\[ E \approx -256.1595061728395 \, \text{N/C} \][/tex]