Two point charges, 4.5 μC and −2 μC are placed at a distance of 0.9 m, the electrostatic potential energy of the system will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −0.09 J
The electrostatic potential energy of two point charges is \( U = q_1 q_2/(4\pi\varepsilon_0 r) \), taken with the signs of the charges; unlike charges give a negative energy.
\( \displaystyle U = \frac{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \times (4.5\times10^{-6}\,\mathrm{C}) \times (-2\times10^{-6}\,\mathrm{C})}{0.9\,\mathrm{m}} \)
\( \displaystyle U = \frac{-8.1\times10^{-2}\,\mathrm{J\,m}}{0.9\,\mathrm{m}} = -0.09\,\mathrm{J} \)