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A parallel plate capacitor has a plate area of 120 cm², a plate separation of 1.8 cm, and a potential difference of 52 V. A dielectric slab of thickness 0.6 mm and dielectric constant 3 is inserted be

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2 parts
a

5.9 pF\text{5.9 pF}

b

3.068×10−10 C\text{3.068} \times 10^{-10} \text{ C}

2.89×103 V/m\text{2.89} \times 10^3 \text{ V/m}

Step-by-step solution

6 steps

12:

First, convert all given values to SI units:
Plate area, A=120 cm2=120×(10−2 m)2=0.012 m2A = 120 \text{ cm}^2 = 120 \times (10^{-2} \text{ m})^2 = 0.012 \text{ m}^2
Plate separation, d=1.8 cm=1.8×10−2 m=0.018 md = 1.8 \text{ cm} = 1.8 \times 10^{-2} \text{ m} = 0.018 \text{ m}
Potential difference, V=52 VV = 52 \text{ V}
Dielectric slab thickness, t=0.6 mm=0.6×10−3 mt = 0.6 \text{ mm} = 0.6 \times 10^{-3} \text{ m}
Dielectric constant, k=3k = 3
Permittivity of free space, ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}

Part a· 2 steps

the capacitance of the capacitor

  1. Use the formula for the capacitance of a parallel plate capacitor.

    C=ϵ0AdC = \frac{\epsilon_0 A}{d}

  2. Substitute the values and calculate the capacitance.

    C=(8.85×10−12 F/m)×(0.012 m2)0.018 mC = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (0.012 \text{ m}^2)}{0.018 \text{ m}} C=1.062×10−13 F⋅m0.018 mC = \frac{1.062 \times 10^{-13} \text{ F}\cdot\text{m}}{0.018 \text{ m}} C=5.9×10−12 FC = 5.9 \times 10^{-12} \text{ F}

    The capacitance of the capacitor is 5.9 pF\boxed{\text{5.9 pF}}.

Part b

3.068×10−10 C\text{3.068} \times 10^{-10} \text{ C}

what free charge appears on the plate

  1. Use the relationship between charge, capacitance, and voltage.

    Q=CVQ = CV

  2. Substitute the capacitance from part (a) and the given voltage.

    Q=(5.9×10−12 F)×(52 V)Q = (5.9 \times 10^{-12} \text{ F}) \times (52 \text{ V}) Q=306.8×10−12 CQ = 306.8 \times 10^{-12} \text{ C} Q=3.068×10−10 CQ = 3.068 \times 10^{-10} \text{ C}

    The free charge on the plate is 3.068×10−10 C\boxed{\text{3.068} \times 10^{-10} \text{ C}}.

    c) If the battery was disconnected and a dielectric slab of thickness 0.6 mm0.6 \text{ mm} was inserted between the plates, what is the electric field in the gap between the plates and the dielectric slab if k=3k=3

  3. When the battery is disconnected, the charge QQ on the plates remains constant.

    The electric field in the air gap (the region without the dielectric) is determined by this constant charge density.
    The electric field in the air gap is given by Egap=QAϵ0E_{gap} = \frac{Q}{A \epsilon_0}. This is equivalent to the initial electric field before the dielectric was inserted, Einitial=VdE_{initial} = \frac{V}{d}.

  4. Calculate the electric field in the gap using the initial voltage and plate separation.

    Egap=VdE_{gap} = \frac{V}{d} Egap=52 V0.018 mE_{gap} = \frac{52 \text{ V}}{0.018 \text{ m}} Egap≈2888.89 V/mE_{gap} \approx 2888.89 \text{ V/m}

    The electric field in the gap between the plates and the dielectric slab is 2.89×103 V/m\boxed{\text{2.89} \times 10^3 \text{ V/m}}.

Part d

what is the electric field in the dielectric slab.

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