A cylindrical capacitor consists of a solid inner conducting core with radius 0.300 cm…?

A cylindrical capacitor consists of a solid inner conducting core with radius 0.300 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 11.5 cm. The capacitance is 37.0 pF.

1) Calculate the outer radius of the hollow tube.

r=???? cm

2)When the capacitor is charged to 120 V, what is the charge per unit length lambda on the capacitor?

λ=??? C/m

5 Answers

  • Capacitance/unit length of a long cylinder is

    C/L = (2πεrε₀) / (ln (b/a))

    b is radius of outside conductor

    a is radius if inside conductor

    ε₀ is 8.8542e-12 F/m

    εᵣ is dielectric constant (vacuum = 1)

    C = 37pF = (0.115)(2π*8.85e-12) / (ln (b/a))

    ln (b/a) = 0.17283

    b/a = 1.189

    b = 1.189×0.3 = 0.357 cm

    Second half, use Q = CV to get charge and divide by 11.5 cm to get charge per cm.

    .

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    I cannot conceive of such an object. “Infinitely long, radius of 2.8 cm and zero thickness???” What universe are we talking of?

  • Capacitance is C = Q/V

    Q = λL

    V = (λ/2πε₀) ln(r₁/r₂)

    So…

    C = (2πε₀L) / ln(r₁/r₂)

    r₁ is radius of outside conductor

    r₂ is radius if inside conductor

    ε₀ is 8.8542e-12 F/m = 8.85 pF

    C = 37pF = 37e-12F=(0.115 m)(2π*8.85e-12 F) / ln (r₁/r₂)

    ln (r₁/r₂) = 0.17283

    then convert this to exponential form

    e^(0.17283) = r₁/r₂

    Solver for r₁ and substitute for r₂

    r₁ = 0.357 cm

    Source(s): Physics textbook.
  • Can anybody answer more clearly

  • q=cv

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