Two charges, $5$ nC at $(2\,\text{cm},0,0)$ and $-2$ nC at $(20\,\text{cm},0,0)$, are placed in free space with no external field. Their mutual electrostatic potential energy is approximately:
A$+5$ μJ
B$-5$ μJ
C$-0.5$ μJ
D$+0.5$ μJ
Answer & Solution
Correct answer: C. $-0.5$ μJ
Separation $r = 18$ cm $= 0.18$ m. $U = (1/4\pi\varepsilon_0)q_1q_2/r = (9\times10^9)(5\times10^{-9})(-2\times10^{-9})/0.18 = -5\times10^{-7}$ J $= -0.5$ μJ. Negative because the charges attract — energy was released when they were brought together.
Related questions
If a positive point charge is placed at the centre of a spherical conducting shell of radiThe SI unit of electric field isFor a charge distribution with volume density ρ, the total charge Q equalsThe electric flux through a closed surface enclosing a charge of +5 μC is (ε₀ = 8.85 × 10⁻Two charges +q and −q of equal magnitude are placed at the corners A and B of an equilaterThe principle of quantisation of electric charge is stated asAn electric dipole in a uniform electric field E experiencesThe electric field on the axial line of a short electric dipole at distance r is