Practice question

Given: Three charges +2 μC, -3 μC, and +1 μC are at (0, 0, 0), (2, 0, 0), and (0, 2, 0) m . What is the potential at (2, 2, 0) m ? (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: Distances: r_1 = sqrt2² + 2² = 2sqrt2, r_2 = 2 m, r_3 = 2 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: V = 9 × 10⁹( frac2 × 10⁻⁶² sqrt2 + frac-3 × 10⁻⁶²+ frac1 × 10⁻⁶²) . V = 9 × 10⁹( frac2 × 10⁻⁶².828 - frac3 × 10⁻⁶²+ frac1 × 10⁻⁶²) . V = 9 × 10⁹( 0.707 × 10⁻⁶- 1.5 × 10⁻⁶+ 0.5 × 10⁻⁶) = 9 × 10⁹ × (-0.293 × 10⁻⁶) = -2637 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.