what is the electric potential 4.20 m from a -3.37*10^-6 c charge? include the sign, + or - . (unit = v)

Explanation:

Step1: Recall electric - potential formula

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

Step2: Identify given values

We are given $q=- 3.37\times10^{-6}\ C$ and $r = 4.20\ m$.

Step3: Substitute values into formula

$V=\frac{(9\times10^{9}\ N\cdot m^{2}/C^{2})\times(-3.37\times10^{-6}\ C)}{4.20\ m}$. First, calculate the numerator: $(9\times10^{9})\times(-3.37\times10^{-6})=-30.33\times10^{3}=-3.033\times10^{4}$. Then, divide by the denominator: $V=\frac{-3.033\times10^{4}}{4.20}\ V\approx - 7221.43\ V$.

Answer:

$-7221\ V$