in the diagram, q1 = -2.00 x 10-5 c, q2 = +3.80 x 10-6 c, and q3 = +5.30 x 10-5 c. what is the electric…

Explanation:

Step1: Recall electric - potential - energy formula

The electric - potential energy between two point charges $q_i$ and $q_j$ separated by a distance $r_{ij}$ is given by $U = k\frac{q_iq_j}{r_{ij}}$, where $k = 9.0\times10^{9}\ N\cdot m^{2}/C^{2}$. The total electric - potential energy of $q_2$ due to $q_1$ and $q_3$ is $U_{e}=U_{12}+U_{23}$, where $U_{12}=k\frac{q_1q_2}{r_{12}}$ and $U_{23}=k\frac{q_2q_3}{r_{23}}$.

Step2: Calculate $U_{12}$

Given $q_1=- 2.00\times10^{-5}\ C$, $q_2 = 3.80\times10^{-6}\ C$, and $r_{12}=1.15\ m$. $U_{12}=k\frac{q_1q_2}{r_{12}}=(9.0\times10^{9}\ N\cdot m^{2}/C^{2})\frac{(-2.00\times10^{-5}\ C)\times(3.80\times10^{-6}\ C)}{1.15\ m}$ $U_{12}=(9.0\times10^{9})\frac{-7.6\times10^{-11}}{1.15}\ J=- 6.0\times10^{-1}\ J$.

Step3: Calculate $U_{23}$

Given $q_2 = 3.80\times10^{-6}\ C$, $q_3 = 5.30\times10^{-5}\ C$, and $r_{23}=2.88\ m$. $U_{23}=k\frac{q_2q_3}{r_{23}}=(9.0\times10^{9}\ N\cdot m^{2}/C^{2})\frac{(3.80\times10^{-6}\ C)\times(5.30\times10^{-5}\ C)}{2.88\ m}$ $U_{23}=(9.0\times10^{9})\frac{2.014\times10^{-10}}{2.88}\ J\approx0.63\ J$.

Step4: Calculate $U_{e}$

$U_{e}=U_{12}+U_{23}=-0.60\ J + 0.63\ J=0.03\ J$.

Answer:

$0.03\ J$