Practice question
Question
Two charges \( 14 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are at \( (3, 0, 0) \) and \( (-3, 0,
0) \, \text{cm} \). What is the potential at midpoint? (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. Distance to midpoint = 0.03 m. V = 9 × 10⁹ ( (14 × 10⁻⁶/0.03) + (-6 × 10⁻⁶/0.03) ) = 9 × 10⁹ × (8 × 10⁻⁶/0.03) . V = 9 × 10⁹ × (8 × 10⁻⁶/0.03) = 2.4 × 10⁶ V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq
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