Practice question

Given: Two charges 4 × 10⁻⁸ C and -2 × 10⁻⁸ C are 20 cm apart. At what distance from the positive charge on the line joining them is the potential zero? (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: Total potential: V = 1/4 π varepsilon_0 ( frac4 × 10⁻⁸ x + frac-2 × 10⁻⁸⁰.2 - x ) = 0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Let distance from 4 × 10⁻⁸ C be x m, then distance from -2 × 10⁻⁸ C is 0.2 - x . . Simplify: 9 × 10⁹( frac4 × 10⁻⁸ x - frac2 × 10⁻⁸⁰.2 - x ) = 0 . 4/x = 2/0.2 - x Rightarrow 4 (0.2 - x) = 2x Rightarrow 0.8 - 4x = 2x Rightarrow 0.8 = 6x Rightarrow x = 0.8/6 = 0.133 m = 13.3 cm . 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.