a 2.55 x 10-4 c charge is held 0.350 m from a second 7.62 x 10-6 c charge. the second one is released. when…

Explanation:

Step1: Recall electric - potential - energy formula

The electric - potential - energy formula between two 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: Calculate the initial electric - potential energy $U_1$

Let $q_1 = 2.55\times 10^{-4}\ C$, $q_2 = 7.62\times 10^{-6}\ C$, and $r_1=0.350\ m$. Then $U_1 = k\frac{q_1q_2}{r_1}=9\times 10^{9}\times\frac{(2.55\times 10^{-4})\times(7.62\times 10^{-6})}{0.350}$. [ \begin{align*} U_1&=9\times 10^{9}\times\frac{2.55\times7.62\times 10^{-10}}{0.350}\ &=9\times\frac{2.55\times7.62\times 10^{-10 + 9}}{0.350}\ &=9\times\frac{19.431\times 10^{-1}}{0.350}\ &=\frac{174.879\times 10^{-1}}{0.350}\ &\approx 49.965\ J \end{align*} ]

Step3: Calculate the final electric - potential energy $U_2$

Let $r_2 = 1.55\ m$. Then $U_2 = k\frac{q_1q_2}{r_2}=9\times 10^{9}\times\frac{(2.55\times 10^{-4})\times(7.62\times 10^{-6})}{1.55}$. [ \begin{align*} U_2&=9\times 10^{9}\times\frac{2.55\times7.62\times 10^{-10}}{1.55}\ &=9\times\frac{19.431\times 10^{-10 + 9}}{1.55}\ &=9\times\frac{1.9431}{1.55}\ &=\frac{17.4879}{1.55}\ &\approx 11.282\ J \end{align*} ]

Step4: Calculate the change in electric - potential energy $\Delta U$

$\Delta U=U_2 - U_1=11.282 - 49.965=- 38.683\ J$. The negative sign indicates that the electric - potential energy has decreased.

Answer:

$-38.7\ J$ (rounded to one decimal place)