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Give step-by-step solution with explanation and final answer:Acylindrical capacitor has an inner conductor of radius 2.9 mum and an outer conductor of radius 4.0 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.1m long. v PartA What is the capacitance per unit length? Express your answer in picofarads per meter. v PartB The potential of the inner conductor relative to that of the outer conductor is 380 mV . Find the charge (magnitude and sign) on the inner conductor. Express your answer with the appropriate units. HB. os @ =| Value | Units [suomi | Request Answer v PartC The potential of the inner conductor relative to that of the outer conductor is 380 mV. Find the charge (magnitude and sign) on the outer conductor. Express your answer with the appropriate units.

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Give step-by-step solution with explanation and final answer:Uploaded ImageAcylindrical capacitor has an inner conductor of radius 2.9 mum and an outer conductor of radius 4.0 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.1m long. v PartA What is the capacitance per unit length? Express your answer in picofarads per meter. v PartB The potential of the inner conductor relative to that of the outer conductor is 380 mV . Find the charge (magnitude and sign) on the inner conductor. Express your answer with the appropriate units. HB. os @ =| Value | Units [suomi | Request Answer v PartC The potential of the inner conductor relative to that of the outer conductor is 380 mV. Find the charge (magnitude and sign) on the outer conductor. Express your answer with the appropriate units.

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Asked by: Rahul
Created at: 2026-02-16 04:12:10

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Let's solve each part step by step: --- ## **Given Data** - Inner conductor radius, \( a = 2.9 \) mm \( = 2.9 \times 10^{-3} \) m - Outer conductor radius, \( b = 4. \) mm \( = 4. \times 10^{-3} \) m - Length of capacitor, \( L = 2.1 \) m - Vacuum between conductors - Potential difference (\( V \)) for parts B & C: \( 380 \) mV \( = .380 \) V --- ## **Part A: Capacitance per unit length** ### **Formula for Cylindrical Capacitor Capacitance per unit length:** \[ C' = \frac{2\pi \varepsilon_}{\ln(b/a)} \] Where: - \( C' \) = capacitance per unit length (F/m) - \( \varepsilon_ = 8.854 \times 10^{-12} \) F/m (vacuum permittivity) - \( a \) = inner radius (m) - \( b \) = outer radius (m) ### **Calculation** 1. Calculate \( \ln(b/a) \): \[ \ln\left(\frac{b}{a}\right) = \ln\left(\frac{4. \times 10^{-3}}{2.9 \times 10^{-3}}\right) = \ln(1.37931) \approx .321 \] 2. Substitute values: \[ C' = \frac{2 \pi (8.854 \times 10^{-12})}{.321} \] \[ = \frac{2 \times 3.1416 \times 8.854 \times 10^{-12}}{.321} \] \[ = \frac{55.614 \times 10^{-12}}{.321} \] \[ = 173.25 \times 10^{-12} \text{ F/m} \] \[ = 173.25 \text{ pF/m} \] **Final Answer Part A:** \[ \boxed{173 \ \text{pF/m}} \] --- ## **Part B: Charge on the Inner Conductor** ### **Formula:** \[ Q = C \cdot V \] But for the total capacitor: \[ C = C' \cdot L \] So: \[ Q = (C' \cdot L) \cdot V \] ### **Calculation:** \[ C' = 173.25 \times 10^{-12} \ \text{F/m} \] \[ L = 2.1 \ \text{m} \] \[ V = .380 \ \text{V} \] \[ C = 173.25 \times 10^{-12} \times 2.1 = 363.825 \times 10^{-12} \ \text{F} \] \[ Q = 363.825 \times 10^{-12} \times .380 = 138.25 \times 10^{-12} \ \text{C} \] \[ Q \approx 1.38 \times 10^{-10} \ \text{C} \] **Final Answer Part B:** \[ \boxed{1.38 \times 10^{-10} \ \text{C}} \] (The charge is **positive** on the inner conductor.) --- ## **Part C: Charge on the Outer Conductor** - By conservation of charge, if inner is \( +Q \), outer is \( -Q \). **Final Answer Part C:** \[ \boxed{-1.38 \times 10^{-10} \ \text{C}} \] (The charge is **negative** on the outer conductor.) --- ### **Summary Table** | Part | Quantity | Value | |------|------------------------------|--------------------------------| | A | Capacitance per unit length | \( 173 \) pF/m | | B | Charge on inner conductor | \( +1.38 \times 10^{-10} \) C | | C | Charge on outer conductor | \( -1.38 \times 10^{-10} \) C | Let me know if you need any clarifications!

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