Chapter 24 — Capacitance & Dielectrics

14 points · 3 extra credit

Chapter 24 — Capacitance & Dielectrics

Core: 14 points · Extra credit: 3 points. Show your reasoning; partial credit is given. Use ε0=8.85×10−12 C2/(N⋅m2)\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2) and k=8.99×109 N⋅m2/C2k=8.99\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2.

Worked solutions are available for every problem.
  1. 1. (2 pts ••) A parallel-plate capacitor has plate area A=0.012 m2A=0.012\,\text{m}^2 and separation d=1.5 mmd=1.5\,\text{mm}, connected to a 12 V12\,\text{V} battery.

    (a) Find the capacitance.

    (b) Find the charge on each plate.

    (c) Find the energy stored.

    (d) Find the electric field between the plates.

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  2. 2. (2 pts ••) Three capacitors have capacitances 6.0 μF6.0\,\mu\text{F}, 3.0 μF3.0\,\mu\text{F}, and 2.0 μF2.0\,\mu\text{F}.

    (a) Find the equivalent capacitance if all three are in series.

    (b) Find the equivalent capacitance if all three are in parallel.

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  3. 3. (3 pts •••) A 6.0 μF6.0\,\mu\text{F} capacitor (C1C_1) is connected in series with the parallel combination of two 3.0 μF3.0\,\mu\text{F} capacitors (C2C_2 and C3C_3). This combination is connected across a 24 V24\,\text{V} battery.

    (a) Find the equivalent capacitance of the network.

    (b) Find the total charge delivered by the battery.

    (c) Find the voltage across, and the charge on, each of the three capacitors.

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  4. 4. (2 pts ••) A 4.0 μF4.0\,\mu\text{F} capacitor is charged to 50 V50\,\text{V} and then disconnected from the battery. Its plate separation is then doubled.

    (a) Find the new capacitance.

    (b) Find the new charge on the plates.

    (c) Find the new voltage.

    (d) Find the change in stored energy, and explain where that energy came from.

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  5. 5. (2 pts ••) An air-filled capacitor with C0=100 pFC_0=100\,\text{pF} is connected to a 12 V12\,\text{V} battery. While it remains connected, a dielectric of constant K=3.0K=3.0 is inserted, completely filling the gap.

    (a) Find the new capacitance.

    (b) Find the new charge on the plates.

    (c) Does the stored energy increase or decrease, and where does the difference come from?

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  6. 6. (1 pt •) A capacitor stores 4.5×10−4 J4.5\times10^{-4}\,\text{J} of energy when charged to 30 V30\,\text{V}. Find its capacitance.

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  7. 7. (2 pts ••) A spherical capacitor has an inner sphere of radius a=1.0 cma=1.0\,\text{cm} and a concentric outer shell of radius b=1.2 cmb=1.2\,\text{cm} (its capacitance is C=4πε0abb−aC=4\pi\varepsilon_0\dfrac{ab}{b-a}).

    (a) Find the capacitance.

    (b) Find the energy stored when it is charged to 500 V500\,\text{V}.

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  8. Extra Credit (3 pts •••)

    A cylindrical capacitor consists of an inner conductor of radius aa and a coaxial outer conductor of radius bb, both of length L≫bL\gg b, carrying charges ±Q\pm Q.

    (a) Use Gauss's law to find the electric field between the conductors.

    (b) Integrate the field to find the potential difference between them, and hence show that C=2πε0Lln⁡(b/a)C=\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}.

    (c) Show that when the gap d=b−ad=b-a is very small compared with aa, this reduces to the parallel-plate result C=ε0A/dC=\varepsilon_0 A/d.

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