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Question

A spherical conductor of radius 8 cm has a charge of \( 4 \times 10^{-8} \, \text{C} \). What is the
electric field at 12 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

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. For r = 0.12 m > R = 0.08 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (4 × 10⁻⁸/(0.12)²) = 9 × 10⁹ × (4 × 10⁻⁸/0.0144) = 2.5 × 10⁴ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C

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