Chapter 17 Electric Potential. Objectives: The students will be able to: Given the dimensions, distance between the plates, and the dielectric constant.

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

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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?