1. Chapter 17 Electric Potential

2. 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.

3. 17.7 Capacitance A capacitor consists of two conductors that are close but not touching. A capacitor has the ability to store electric charge.

4. Capacitor Named for the capacity to store electric charge and energy. A capacitor is two conducting plates separated by a finite distance:

5. 17.7 Capacitance Parallel-plate capacitor connected to battery. (b) is a circuit diagram.battery

6. 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

7. Figure 17-15 Key of a computer keyboard

8. 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?

9. 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? V = 16 V, C = 0.75  F = 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

10. Q = CV (17 - 7)

11. 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 ε o is the permittivity of free space, which, as we saw in Chapter 16, has the value 8.85 x 10 –12 C 2 N  m 2.

12. 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:

13. 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.

14. Capacitor Geometry The capacitance of a capacitor depends on HOW you make it.

15. 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,    = 8.85 x 10 -12 C 2 /(Nm 2 )

16. 1.13 x 10 8 m 2 10629 m Is this a practical capacitor to build? NO! – How can you build this then? 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? Add a DIELECTRIC!!!

17. Dielectric A dielectric material (dielectric for short) is an electrical insulator that can be polarized by an applied electric field.

18. 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).

19. 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).

20. 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.

21. Example 17-8 Capacitor calculations. (b) What is the charge on each plate if a 12-V battery is connected across the two plates?

22. Example 17-8 Capacitor calculations. (c) What is the electric field between the plates?

23. 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).

24. Elaboration Capacitors - pHET

25. Homework Problems in chapter 7 31, 34, 37, 38, 40

26. Closure When a battery is connected to a capacitor, why do the plates acquire charges of the same magnitude?

27. Closure When a battery is connected to a capacitor, why do the plates acquire charges of the same magnitude?