Question

Three equal 1.20 µC point charges are placed at the corners of an equilateral triangle with sides 0.400 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Short answer

The potential energy of the system is 0.097 J_.

Step-by-step solution

Potential energy formula

The potential energy of a body is given by

$\mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\mathbf{\sum }_{\mathbf{i}\mathbf{<}\mathbf{j}}\frac{{\mathbf{q}}_{\mathbf{i}}{\mathbf{q}}_{\mathbf{j}}}{{\mathbf{r}}_{\mathbf{ij}}}$

Consider the three charges arethe same so the potential energy will be as follows:

$$\begin{gathered} U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\sum _{I<J}\frac{q_i{q}_j}{r_{ij}} \\ U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q_i{q}_j}{r_{12}}+\frac{q_2{q}_3}{r_{23}}+\frac{q_1{q}_3}{r_{13}}\right) \\ \mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\frac{\mathbf{3}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{r}} \end{gathered}$$

Here,${\epsilon }_0$ is the permittivity of the free space.

Here, $\frac{1}{4{\mathrm{\pi \epsilon }}_0}=9\times {10}^9\frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}$.

Determine the potential energy of the system

Substitute the values and solve as:

$$\begin{gathered} \mathrm{U}=\left(9\times {10}^9\frac{\mathrm{N}.{\mathrm{m}}^2}{{\mathrm{C}}^2}\right)\left(\frac{3{\left(1.20\times {10}^{-6}\mathrm{C}\right)}^2}{0.4\mathrm{m}}\right) \\ =0.097\mathrm{J} \end{gathered}$$

Therefore, the potential energy of the system is 0.097 J