a 2.55 x 10-4 c charge has an electric potential energy of 35.7 j when it is placed 0.437 m from a second…

Explanation:

Step1: Recall electric - potential - energy formula

The electric - potential energy between two charges is given by $U = k\frac{q_1q_2}{r}$, where $U$ is the electric - potential energy, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. We want to solve for $q_2$. Rearranging the formula gives $q_2=\frac{Ur}{kq_1}$.

Step2: Substitute the given values

We are given that $U = 35.7\ J$, $r = 0.437\ m$, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, and $q_1=2.55\times 10^{-4}\ C$. [ \begin{align*} q_2&=\frac{35.7\times0.437}{9\times 10^{9}\times2.55\times 10^{-4}}\ &=\frac{35.7\times0.437}{9\times2.55\times 10^{5}}\ &=\frac{15.6}{22.95\times 10^{5}}\ &=\frac{15.6}{2.295\times 10^{6}}\ & = 6.8\times10^{-6}\ C \end{align*} ]

Answer:

6.8