Practice question

**Sign of potential energy** indicates nature of interaction, like charges U>0, opposite U<0. Work done by field moving positive charge from higher to lower potential positive, as field does work W = q ΔV, ΔV negative when moving to lower V. For two opposite charges +q and -q , the potential energy is U = (1/4 π ε₀) (q₁ q₂/r) = -(1/4 π ε₀) (q²/r) , negative due to attraction. Decreasing r makes U more negative (e.g., from -1/r₁ to -1/r₂ , r₂ < r₁ , U decreases). However, the question interprets "increase" as increasing in magnitude (more negative), as the attractive energy becomes stronger,