To determine the electric field at a point 0.450 meters to the left of a charge [tex]\( q = -5.77 \times 10^{-9} \)[/tex] C, we follow these steps:
1. Identify and Write Down the Known Values:
- Charge [tex]\( q = -5.77 \times 10^{-9} \)[/tex] C
- Distance [tex]\( r = 0.450 \)[/tex] m
- Coulomb’s constant [tex]\( k = 8.99 \times 10^9 \)[/tex] N·m²/C²
2. Formula for the Electric Field:
The electric field [tex]\( E \)[/tex] due to a point charge is given by:
[tex]\[
E = \frac{k \cdot |q|}{r^2}
\][/tex]
Here, [tex]\( |q| \)[/tex] represents the magnitude of the charge.
3. Calculate the Magnitude:
[tex]\[
E = \frac{8.99 \times 10^9 \times 5.77 \times 10^{-9}}{(0.450)^2}
\][/tex]
4. Plug in the Numbers and Solve:
[tex]\[
E = \frac{8.99 \times 10^9 \times 5.77 \times 10^{-9}}{0.2025}
\][/tex]
[tex]\[
E = \frac{51.8723 \times 10^0}{0.2025}
\][/tex]
[tex]\[
E = 256.1595061728395 \, \text{N/C}
\][/tex]
This gives the magnitude of the electric field.
5. Determine the Direction:
The direction of the electric field due to a negative charge is towards the charge. Since the point in question is to the left of the charge, the electric field points to the right, towards the charge.
Therefore, the electric field at the point 0.450 meters to the left of the charge is:
[tex]\[
E = -256.1595061728395 \, \text{N/C}
\][/tex]
The negative sign indicates that the field is directed towards the charge, consistent with the direction indicated.
In conclusion, the electric field at a point 0.450 meters to the left of a [tex]\(-5.77 \times 10^{-9} \)[/tex] C charge is [tex]\(-256.1595061728395 \, \text{N/C}\)[/tex], where the negative sign indicates the direction is towards the charge.
