Question

The electric potential energy of the system of the charges is \[\left[\mathrm{K}=\frac{1}{4\pi\in_{0}}\right](\in_{0}\]= permittivity of free space)

Options

• \[\frac{31\mathrm{Kq}}{\mathrm{r}}\]

• \[\frac{30\mathrm{Kq}}{\mathrm{r}}\]

• \[\frac{29\mathrm{Kq}}{\mathrm{r}}\]

• \[\frac{28\mathrm{Kq}}{\mathrm{r}}\]

Answer 1

\[\frac{31\mathrm{Kq}}{\mathrm{r}}\]

Explanation:

Potential energy of a system of three point charges is given by

\[\mathrm{U}=\frac{1}{4\pi\varepsilon_0}\left(\frac{\mathrm{q}_1\mathrm{q}_2}{\mathrm{r}_{12}}+\frac{\mathrm{q}_2\mathrm{q}_3}{\mathrm{r}_{23}}+\frac{\mathrm{q}_2\mathrm{q}_1}{\mathrm{r}_{31}}\right)\]

\[\mathrm{U}=\mathrm{K}\left(\frac{15\mathrm{q}\times8\mathrm{q}}{5\mathrm{r}}+\frac{8\mathrm{q}\times\mathrm{q}}{4\mathrm{r}}+\frac{\mathrm{q}\times15\mathrm{q}}{3\mathrm{r}}\right)\]

\[\mathrm{U=\frac{31Kq^2}{r}}\]