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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

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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

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?

A parallel plate capacitor is connected to a battery. As the plates are moved farther apart, what happens to each of the following? a) the potential difference across the plates b) the charge on the plates c) the electric field between the plates

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.

A \(4.00 \cdot 10^{3}-n F\) parallel plate capacitor is connected to a \(12.0-\mathrm{V}\) battery and charged. a) What is the charge \(Q\) on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor? The \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor is then disconnected from the \(12.0-\mathrm{V}\) battery and used to charge three uncharged capacitors, a \(100 .-n F\) capacitor, a \(200 .-\mathrm{nF}\) capacitor, and a \(300 .-\mathrm{nF}\) capacitor, connected in series. c) After charging, what is the potential difference across each of the four capacitors? d) How much of the electrical energy stored in the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor was transferred to the other three capacitors?

The distance between the plates of a parallel plate capacitor is reduced by half and the area of the plates is doubled. What happens to the capacitance? a) It remains unchanged. b) It doubles. c) It quadruples. d) It is reduced by half.
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