Three point charges $\displaystyle \mathrm{Q}_{1}(-15 \mu \mathrm{C}), \mathrm{Q}_{2}(10 \mu \mathrm{C})$ and $\displaystyle \mathrm{Q}_{3}(16 \mu \mathrm{C})$ are located at $\displaystyle (0 \mathrm{~cm}, 0 \mathrm{~cm}),(0 \mathrm{~cm}, 3 \mathrm{~cm})$ and $\displaystyle (4 \mathrm{~cm}, 3 \mathrm{~cm})$ respectively. Calculate the electrostatic potential energy of this system of charges.

\[\begin{aligned} & =-43.2 \mathrm{~J} \\ \mathrm{U}_{\text {net }} & =\mathrm{U}_{\mathrm{Q} 1 \mathrm{Q} 2}+\mathrm{U}_{\mathrm{Q} 2 \mathrm{Q} 3}+\mathrm{U}_{\mathrm{Q} 1 \mathrm{Q} 3} \\ & =-45+36-43.2 \\ & =-52.2 \mathrm{~J} \end{aligned} \] Alternatively \[\begin{aligned} & \mathrm{U}=k\left(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{r_{12}}+\frac{\mathrm{Q}_{2} \mathrm{Q}_{3}}{r_{23}}+\frac{\mathrm{Q}_{1} \mathrm{Q}_{3}}{r_{13}}\right) \\ & =\frac{9 \times 10^{9} \times\left(-15 \times 10^{-6}\right) \times\left(10 \times 10^{-6}\right)}{3 \times 10^{-2}}+\frac{9 \times 10^{9} \times\left(10 \times 10^{-6}\right) \times\left(16 \times 10^{-6}\right)}{4 \times 10^{-2}}+\frac{9 \times 10^{9} \times\left(-15 \times 10^{-6}\right) \times\left(16 \times 10^{-6}\right)}{5 \times 10^{-2}} \\ & =(-45+36-43.2) \mathrm{J} \\ & =52.2 \mathrm{~J} \end{aligned} \]