what is the electric field at a point 0.200 m to the right of a +1.50*10^-8 c charge? include a + or - sign…

Explanation:

Step1: Recall electric - field formula

The electric - field formula due to a point charge is $E = k\frac{q}{r^{2}}$, where $k = 9.0\times10^{9}\ N\cdot m^{2}/C^{2}$, $q$ is the charge, and $r$ is the distance from the charge.

Step2: Identify given values

$q = 1.50\times10^{-8}\ C$, $r = 0.200\ m$.

Step3: Calculate the electric - field magnitude

$E=(9.0\times10^{9}\ N\cdot m^{2}/C^{2})\frac{1.50\times10^{-8}\ C}{(0.200\ m)^{2}}$. First, calculate the denominator: $(0.200\ m)^{2}=0.04\ m^{2}$. Then, calculate the product in the numerator: $9.0\times10^{9}\times1.50\times10^{-8}=135\ N\cdot m^{2}/C$. Finally, $E=\frac{135\ N\cdot m^{2}/C}{0.04\ m^{2}} = 3375\ N/C$.

Step4: Determine the direction

Since the charge is positive, the electric - field points away from the charge. If we consider the right - hand direction as positive, the sign of the electric field is +.

Answer:

$+ 3375\ N/C$