Practice question
Question
Two charges \( 12 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are at \( (5, 0, 0) \) and \( (-5, 0,
0) \, \text{cm} \). What is the potential at midpoint? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times
10^9 \, \text{Nm}^2 \text{C}^{-2} \)).
Explanation
**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. Distance to midpoint = 0.05 m. V = 9 × 10⁹ ( (12 × 10⁻⁶/0.05) + (-3 × 10⁻⁶/0.05) ) = 9 × 10⁹ × (9 × 10⁻⁶/0.05) . V = 9 × 10⁹ × (9 × 10⁻⁶/0.05) = 1.62 × 10⁶ 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
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