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Chapter 17 Electric Potential Objectives: The students will be able to: • Given the dimensions, distance between the plates, and the dielectric constant of the material between the plates, determine the magnitude of the capacitance of a parallel plate capacitor. • Given the capacitance, the dielectric constant, and either the potential difference or the charge stored on the plates of a parallel plate capacitor, determine the energy and the energy density stored in the capacitor. 17.7 Capacitance A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge. Capacitor • Named for the capacity to store electric charge and energy. • A capacitor is two conducting plates separated by a finite distance: 17.7 Capacitance Parallel-plate capacitor connected to battery. (b) is a circuit diagram. 17.7 Capacitance When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage: (17-7) The quantity C is called the capacitance. Unit of capacitance: the farad (F) 1 F = 1 C/V Figure 17-15 Key of a computer keyboard Sample Problem: A 0.75 F capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor? Sample Problem: A 0.75 mF capacitor is charged to a voltage of 16 volts. What is the magnitude of the charge on each plate of the capacitor? V = 16 V, C = 0.75 mF = 0.75 x 10-6 F Q =? C = Q/V or Q = CV Q = (0.75 x 10-6)(16) Q = 1.2 x 10-5 C Q = CV (17 - 7) 17.7 Capacitance The capacitance does not depend on the voltage; it is a function of the geometry and materials of the capacitor. For a parallel-plate capacitor: (17-8) We see that C depends only on geometric factors, A and d , and not on Q or V . We derive this useful relation in the optional subsection at the end of this Section. The constant εo is the permittivity of free space , which, as we saw in Chapter 16, has the value 8.85 x 10–12 C2Nm2. A simple type of capacitor is the parallel-plate capacitor. It consists of two plates of area A separated by a distance d. By calculating the electric field created by the charges ±Q, we find that the capacitance of a parallel-plate capacitor is: The general properties of a parallel-plate capacitor – that the capacitance increases as the plates become larger and decreases as the separation increases – are common to all capacitors. Capacitor Geometry The capacitance of a capacitor depends on HOW you make it. 1 C A C d A area of plate d distance beteween plates A C d o constant of proportion ality o vacuum permittivi ty constant o 8.85 x10 12 C2 Nm 2 o A C d • Sample Problem: What is the AREA of a 1 F capacitor that has a plate separation of 1 mm? C = 1 F, d = 1 mm = 0.001 m, o = 8.85 x 10-12 C2/(Nm2) Is this a practical capacitor to build? NO! – How can you build this then? A C o D A 1 8.85 x10 0.001 A 1.13 x 108 m2 The answer lies in REDUCING the AREA. But you must have a CAPACITANCE of 1 F. How can you keep the capacitance at 1 F and reduce the Area at the same time? Sides 10629 m Add a DIELECTRIC!!! 12 Dielectric A dielectric material (dielectric for short) is an electrical insulator that can be polarized by an applied electric field. Example 17-8 Capacitor calculations. (a)Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm x 3.0 cm and are separated by a 1.0-mm air gap. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, assuming the air gap d is 100 times smaller, or 10 microns (1 micron = 1 μm = 10-6 m). Example 17-8 Capacitor calculations. (a)Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm x 3.0 cm and are separated by a 1.0-mm air gap. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, assuming the air gap d is 100 times smaller, or 10 microns (1 micron = 1 μm = 10-6 m). Example 17-8 Capacitor calculations. (a)Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm x 3.0 cm and are separated by a 1.0-mm air gap. Example 17-8 Capacitor calculations. (b) What is the charge on each plate if a 12-V battery is connected across the two plates? Example 17-8 Capacitor calculations. (c) What is the electric field between the plates? Example 17-8 Capacitor calculations. (d) Estimate the area of the plates needed to achieve a capacitance of 1 F, assuming the air gap d is 100 times smaller, or 10 microns (1 micron = 1 μm = 10-6 m). Elaboration • Capacitors - pHET Homework • Problems in chapter 7 • 31, 34, 37, 38, 40 Closure • When a battery is connected to a capacitor, why do the plates acquire charges of the same magnitude? Closure • When a battery is connected to a capacitor, why do the plates acquire charges of the same magnitude?