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Question

A point charge \( Q = 9 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential
at a point 3 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,
\text{Nm}^2 \text{C}^{-2} \)).

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Explanation

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (9 × 10⁻⁹/3) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 V . 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 V² and common potential V =

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