what is the electric field at a point 0.450 m to the left of a -5.77*10^-9 c charge? include a + or - sign…

Explanation:

Step1: Identify the formula

The formula for the electric - field due to a point charge is $E = k\frac{q}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q$ is the charge, and $r$ is the distance from the charge.

Step2: Substitute the values

Given $q=- 5.77\times10^{-9}\ C$ and $r = 0.450\ m$. $E=(9\times10^{9}\ N\cdot m^{2}/C^{2})\frac{-5.77\times10^{-9}\ C}{(0.450\ m)^{2}}$

Step3: Calculate the value

First, calculate the denominator: $(0.450)^{2}=0.2025\ m^{2}$. Then, calculate the product in the numerator: $9\times10^{9}\times(-5.77\times10^{-9})=-51.93\ N\cdot m^{2}/C$. Now, divide: $E=\frac{-51.93\ N\cdot m^{2}/C}{0.2025\ m^{2}}=-256.44\ N/C$. Since the electric field points towards the negatively - charged particle and the point is to the left of the negative charge, the direction of the electric field is to the right, so we take the positive value.

Answer:

$256\ N/C$