1.;A capacitor has a capacitance of 8.0 microFarad with air between its plates. Determine its capacitance when a dielectric with dielectric constant 6.0 is placed between its plates. 2. What is the charge on a 300 pF capacitor when it is charged to a voltage of 1.0 kV? 3. A certain parallel-plate capacitor consists of two plates, each with area 200 cm² , separated by a 0.40 cm air gap. (a) Compute its capacitance. (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates. (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source?

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1. A capacitor has a capacitance of 8.0 microFarad with air between its plates. Determine its capacitance when a dielectric with dielectric constant 6.0 is placed between its plates.

  1. Identify the given values.

    Initial capacitance with air, C0=8.0 µFC_0 = 8.0 \text{ µF}.
    Dielectric constant, K=6.0K = 6.0.

  2. Use the formula for capacitance with a dielectric.

    The capacitance CC with a dielectric is given by C=KC0C = K C_0.

  3. Substitute the values and calculate the new capacitance.

    C=(6.0)×(8.0 µF)C = (6.0) \times (8.0 \text{ µF}) C=48.0 µFC = 48.0 \text{ µF}

    The capacitance with the dielectric is 48.0 µF\boxed{48.0 \text{ µF}}.

    2. What is the charge on a 300 pF capacitor when it is charged to a voltage of 1.0 kV?

  4. Identify the given values and convert units to SI units.

    Capacitance, C=300 pF=300×10−12 FC = 300 \text{ pF} = 300 \times 10^{-12} \text{ F}.
    Voltage, V=1.0 kV=1.0×103 VV = 1.0 \text{ kV} = 1.0 \times 10^3 \text{ V}.

  5. Use the formula relating charge, capacitance, and voltage.

    The charge QQ on a capacitor is given by Q=CVQ = CV.

  6. Substitute the values and calculate the charge.

    Q=(300×10−12 F)×(1.0×103 V)Q = (300 \times 10^{-12} \text{ F}) \times (1.0 \times 10^3 \text{ V}) Q=300×10−9 CQ = 300 \times 10^{-9} \text{ C} Q=3.0×10−7 CQ = 3.0 \times 10^{-7} \text{ C}

    The charge on the capacitor is 3.0×10−7 C\boxed{3.0 \times 10^{-7} \text{ C}}.

    3. A certain parallel-plate capacitor consists of two plates, each with area 200 cm² , separated by a 0.40 cm air gap.

    (a) Compute its capacitance.

    (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates.

    (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source?

    Given values:
    Area, A=200 cm2=200×(10−2 m)2=200×10−4 m2=0.0200 m2A = 200 \text{ cm}^2 = 200 \times (10^{-2} \text{ m})^2 = 200 \times 10^{-4} \text{ m}^2 = 0.0200 \text{ m}^2.
    Separation, d=0.40 cm=0.40×10−2 m=0.0040 md = 0.40 \text{ cm} = 0.40 \times 10^{-2} \text{ m} = 0.0040 \text{ m}.
    Permittivity of free space, ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}.
    Voltage, V=500 VV = 500 \text{ V}.
    Dielectric constant, K=2.60K = 2.60.

    (a) Compute its capacitance.

  7. Use the formula for the capacitance of a parallel-plate capacitor with air.

    C0=ϵ0AdC_0 = \frac{\epsilon_0 A}{d}
  8. Substitute the values and calculate the capacitance.

    C0=(8.85×10−12 F/m)×(0.0200 m2)0.0040 mC_0 = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (0.0200 \text{ m}^2)}{0.0040 \text{ m}} C0=1.77×10−13 F⋅m0.0040 mC_0 = \frac{1.77 \times 10^{-13} \text{ F}\cdot\text{m}}{0.0040 \text{ m}} C0=4.425×10−11 FC_0 = 4.425 \times 10^{-11} \text{ F} C0=44.25 pFC_0 = 44.25 \text{ pF}

    The capacitance is 4.43×10−11 F\boxed{4.43 \times 10^{-11} \text{ F}} (or 44.3 pF44.3 \text{ pF}).

    (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates.

  9. Calculate the charge Q0Q_0 on the capacitor.

    Q0=C0VQ_0 = C_0 V Q0=(4.425×10−11 F)×(500 V)Q_0 = (4.425 \times 10^{-11} \text{ F}) \times (500 \text{ V}) Q0=2.2125×10−8 CQ_0 = 2.2125 \times 10^{-8} \text{ C}

    The charge on the capacitor is 2.21×10−8 C\boxed{2.21 \times 10^{-8} \text{ C}}.

  10. Calculate the energy stored U0U_0 in the capacitor.

    U0=12C0V2U_0 = \frac{1}{2} C_0 V^2 U0=12(4.425×10−11 F)×(500 V)2U_0 = \frac{1}{2} (4.425 \times 10^{-11} \text{ F}) \times (500 \text{ V})^2 U0=12(4.425×10−11 F)×(250000 V2)U_0 = \frac{1}{2} (4.425 \times 10^{-11} \text{ F}) \times (250000 \text{ V}^2) U0=5.53125×10−6 JU_0 = 5.53125 \times 10^{-6} \text{ J}

    The energy stored in the capacitor is 5.53×10−6 J\boxed{5.53 \times 10^{-6} \text{ J}}.

  11. Calculate the electric field E0E_0 between the plates.

    E0=VdE_0 = \frac{V}{d} E0=500 V0.0040 mE_0 = \frac{500 \text{ V}}{0.0040 \text{ m}} E0=125000 V/mE_0 = 125000 \text{ V/m}

    The electric field between the plates is 1.25×105 V/m\boxed{1.25 \times 10^5 \text{ V/m}}.

    (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source?

  12. Calculate the new capacitance CC with the dielectric.

    C=KC0C = K C_0 C=(2.60)×(4.425×10−11 F)C = (2.60) \times (4.425 \times 10^{-11} \text{ F}) C=1.1505×10−10 FC = 1.1505 \times 10^{-10} \text{ F}
  13. Calculate the new charge QQ on the capacitor with the dielectric.

    The voltage source remains V=500 VV = 500 \text{ V}.

    Q=CVQ = C V Q=(1.1505×10−10 F)×(500 V)Q = (1.1505 \times 10^{-10} \text{ F}) \times (500 \text{ V}) Q=5.7525×10−8 CQ = 5.7525 \times 10^{-8} \text{ C}
  14. Calculate the additional charge ΔQ\Delta Q that flows onto the capacitor.

    ΔQ=Q−Q0\Delta Q = Q - Q_0 ΔQ=(5.7525×10−8 C)−(2.2125×10−8 C)\Delta Q = (5.7525 \times 10^{-8} \text{ C}) - (2.2125 \times 10^{-8} \text{ C}) ΔQ=3.54×10−8 C\Delta Q = 3.54 \times 10^{-8} \text{ C}

    The additional charge that will flow onto the capacitor is 3.54×10−8 C\boxed{3.54 \times 10^{-8} \text{ C}}.

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