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Question

A spherical conductor of radius 30 cm has a charge of \( 12 \times 10^{-8} \, \text{C} \). What is the
electric field at 70 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,
\text{Nm}^2 \text{C}^{-2} \)).

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Explanation

**Electric potential** V = k Q/r, k = 1/(4π ε₀)=9×10⁹ N·m²/C², Q charge (C), r distance (m), scalar, potential at surface of spherical conductor radius R, V = k Q/R. Potential difference ΔV = V_B - V_A = -∫ E·dl, work per unit charge to move charge without acceleration. For r = 0.7 m > R = 0.3 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (12 × 10⁻⁸/(0.7)²) = 9 × 10⁹ × (12 × 10⁻⁸/0.49) ≈ 2.204 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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