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Chapter 24: Problem 66

A 4.00 -pF parallel plate capacitor has a potential difference of \(10.0 \mathrm{~V}\) across it. The plates are \(3.00 \mathrm{~mm}\) apart, and the space between them contains air. a) What is the charge on the capacitor? b) How much energy is stored in the capacitor? c) What is the area of the plates? d) What would the capacitance of this capacitor be if the space between the plates were filled with polystyrene?

Short Answer

Expert verified
Question: Calculate the charge, stored energy, area of the plates, and the capacitance of a parallel plate capacitor when it has a capacitance of 4.00 pF, a potential difference of 10.0 V, a distance between the plates of 3.00 mm, and if the space between the plates contains polystyrene material. Answer: The charge on the capacitor is \(4.00 \times 10^{-11}\) coulombs, the energy stored in the capacitor is \(2.00 \times 10^{-10}\) joules, the area of the plates is \(1.18 \times 10^{-6}\) square meters, and the capacitance with polystyrene is \(10.4 \text{ pF}\).

Step by step solution

01

a) Charge on the capacitor

To find the charge, we use the formula \(Q = CV\). We know the capacitance, \(C\), is \(4.00\) pF and the potential difference, \(V\), is \(10.0\) V. First, we have to convert the capacitance to Farads: \(4.00\) pF = \(4.00 \times 10^{-12}\) F. Now we can calculate the charge: $$Q = (4.00 \times 10^{-12}\text{ F})(10.0\text{ V}) = 4.00 \times 10^{-11}\text{ C}$$ The charge on the capacitor is \(4.00 \times 10^{-11}\) coulombs.
02

b) Energy stored in the capacitor

To find the stored energy, we use the formula \(U=\frac{1}{2}CV^2\). We have the capacitance, \(C\), as \(4.00 \times 10^{-12} \text{ F}\) and the potential difference, \(V\), as \(10.0 \text{ V}\). $$U = \frac{1}{2}(4.00 \times 10^{-12}\text{ F})(10.0\text{ V})^2 = 2.00 \times 10^{-10}\text{ J}$$ The energy stored in the capacitor is \(2.00 \times 10^{-10}\) joules.
03

c) Area of the plates

To find the area of the plates, we use the formula \(A = \frac{C\epsilon_0}{V}d\). We have the capacitance, \(C\), as \(4.00 \times 10^{-12} \text{ F}\), the distance between the plates, \(d\), is \(3.00 \mathrm{~mm} = 3.00 \times 10^{-3}\mathrm{~m}\), and the potential difference, \(V\), is \(10.0 \text{ V}\). The vacuum permittivity constant, \(\epsilon_0\), is \(8.85 \times 10^{-12} \text{F}\cdot\text{m}^{-1}\). $$A = \frac{(4.00 \times 10^{-12}\text{ F})(8.85 \times 10^{-12}\text{ F}\cdot\text{m}^{-1})}{(10.0\text{ V})(3.00 \times 10^{-3}\mathrm{~m})} = 1.18 \times 10^{-6} \text{m}^2$$ The area of the plates is \(1.18 \times 10^{-6}\) square meters.
04

d) Capacitance with polystyrene

To find the new capacitance when the space contains polystyrene, we use the formula \(C_2 = KC_1\). We have the initial capacitance, \(C_1\), as \(4.00 \times 10^{-12} \text{ F}\), and the dielectric constant of polystyrene, \(K\), is approximately \(2.6\). $$C_2 = (2.6)(4.00 \times 10^{-12}\text{ F}) = 10.4 \times 10^{-12}\text{ F}$$ The capacitance of this capacitor with polystyrene is \(10.4 \text{ pF}\).

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Most popular questions from this chapter

A \(1.00-\mu \mathrm{F}\) capacitor charged to \(50.0 \mathrm{~V}\) and a \(2.00-\mu \mathrm{F}\) capacitor charged to \(20.0 \mathrm{~V}\) are connected, with the positive plate of each connected to the negative plate of the other. What is the final charge on the \(1.00-\mu \mathrm{F}\) capacitor?

Which of the following is proportional to the capacitance of a parallel plate capacitor? a) the charge stored on each conducting plate b) the potential difference between the two plates c) the separation distance between the two plates d) the area of each plate e) all of the above f) none of the above

A parallel plate capacitor is constructed from two plates of different areas. If this capacitor is initially uncharged and then connected to a battery, how will the amount of charge on the big plate compare to the amount of charge on the small plate?

A parallel plate capacitor with square plates of edge length \(L\) separated by a distance \(d\) is given a charge \(Q\), then disconnected from its power source. A close-fitting square slab of dielectric, with dielectric constant \(\kappa\), is then inserted into the previously empty space between the plates. Calculate the force with which the slab is pulled into the capacitor during the insertion process.

How much energy can be stored in a capacitor with two parallel plates, each with an area of \(64.0 \mathrm{~cm}^{2}\) and separated by a gap of \(1.30 \mathrm{~mm}\), filled with porcelain whose dielectric constant is \(7.0,\) and holding equal and opposite charges of magnitude \(420 . \mu C ?\)
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