Practice question
Question
A point charge \( Q = 12 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential
at a point 6 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,
\text{Nm}^2 \text{C}^{-2} \)).
Explanation
**Equipotential surface** is surface where V constant, no work done moving charge along it because W = q ΔV =0 when ΔV=0. Electric field E always perpendicular to equipotential surface, direction from higher to lower potential, magnitude E = -dV/dr, steeper potential gradient means stronger field. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (12 × 10⁻⁹/6) = 9 × 10⁹ × 2 × 10⁻⁹ = 18 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
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