a 0.00287 c charge is 6.52 m from a -0.00555 c charge. what is their electric potential energy? include the…

Explanation:

Step1: Recall electric - potential - energy formula

The formula for the electric potential energy between two point - charges is $U = k\frac{q_1q_2}{r}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: Identify the values of $q_1$, $q_2$, and $r$

$q_1 = 0.00287\ C$, $q_2=- 0.00555\ C$, and $r = 6.52\ m$.

Step3: Substitute the values into the formula

$U=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\frac{(0.00287\ C)\times(-0.00555\ C)}{6.52\ m}$. First, calculate the product of the charges: $(0.00287\ C)\times(-0.00555\ C)=- 1.59285\times10^{-5}\ C^{2}$. Then, calculate the fraction: $\frac{-1.59285\times 10^{-5}\ C^{2}}{6.52\ m}=-2.443\times10^{-6}\ C^{2}/m$. Finally, multiply by $k$: $U=(9\times10^{9}\ N\cdot m^{2}/C^{2})\times(-2.443\times10^{-6}\ C^{2}/m)=-21987\ J$.

Answer:

$-21987\ J$