Practice question
Question
Three charges \( +5 \, \mu\text{C} \), \( -2 \, \mu\text{C} \), and \( +3 \, \mu\text{C} \) are at \(
(0, 0, 0) \), \( (5, 0, 0) \), and \( (0, 5, 0) \, \text{m} \). What is the potential at \( (5, 5, 0) \,
\text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).
Explanation
**Potential due to point charge** V = (1/4π ε₀)Q/r, positive Q gives positive V, negative gives negative. At r=0.3 m from Q=3×10⁻⁸ C, V=9×10⁹×3×10⁻⁸/0.3=900 V. Work done moving charge q from infinity (V=0) to point where V=40 V is W = q V =9×10⁻⁶×40=3.6×10⁻⁴ J. Distances: r₁ = √(5² + 5²) = 5√(2) m , r₂ = 5 m , r₃ = 5 m . V = 9 × 10⁹ ( (5 × 10⁻⁶/5√(2)) + (-2 × 10⁻⁶/5) + (3 × 10⁻⁶/5) ) . V = 9 × 10⁹ ( (5 × 10⁻⁶/7.07) - (2 × 10⁻⁶/5) + (3 × 10⁻⁶/5) ) . V =
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