Chapter 17 Electric Potential. Objectives: The students will be able to: Given the dimensions,...
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Transcript of Chapter 17 Electric Potential. Objectives: The students will be able to: Given the dimensions,...
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 17-15Key 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 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 = CVQ = (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 GeometryThe capacitance of a
capacitor depends on HOW you make it.
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• 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 C2/(Nm2)
Sides
A
Ax
D
AC o
001.01085.81 12
1.13 x 108 m2
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!!!
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?