Practice question

Given: A dipole with p = 5 × 10⁻¹⁰ C m is along the x-axis. 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: r = sqrt2² + 2² = 2sqrt2 m, cos θ = frac22sqrt2 = frac1sqrt2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: V = 9 × 10⁹ × frac5 × 10⁻¹⁰ × frac1sqrt2(2sqrt2)² = 9 × 10⁹ × frac5 × 10⁻¹⁰⁸ sqrt2 × frac1sqrt2 = 9 × 10⁹ × frac5 × 10⁻¹⁰¹⁶= 2.8125 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.