i chapter 24 capacitance, dielectrics, electric energy storage

18
I Chapter 24 Capacitance, Dielectrics, Electric Energy Storage

Upload: ethelbert-baker

Post on 24-Dec-2015

230 views

Category:

Documents


4 download

TRANSCRIPT

IChapter 24Capacitance, Dielectrics, Electric Energy Storage

I

An equipotential is a line or surface over which the potential is constant.

Electric field lines are perpendicular to equipotentials.

The surface of a conductor is an equipotential.

23-5 Equipotential Surfaces

I 23-5 Equipotential Surfaces

Example 23-10: Point charge equipotential surfaces.

For a single point charge with Q = 4.0 × 10-9 C, sketch the equipotential surfaces (or lines in a plane containing the charge) corresponding to V1 = 10 V, V2 = 20 V, and V3 = 30 V.

I 23-5 Equipotential Surfaces

I 23-5 Equipotential Surfaces

Equipotential surfaces are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges).

I Equipotential Surfaces

Surfaces at same potential like contours on topographic maps

Lines link places at same elevation (same

Ug)

Lines link places at same potential (same

V)

I Equipotential Surfaces

Surfaces that have the same potential (voltage) at every pointElectric Field lines are perpendicular to equipotential surfaces

Potential difference between surfaces is constant

Surfaces are closer where the potential is stronger

Electric field always points in the direction of maximum potential DECREASE

I QuestionWhich image below best shows the equipotential and Electric field lines

+-+-

+- +-

A B

C D

I

The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation:

23-6 Electric Dipole Potential

I23-7 E Determined from V

If we know the field, we can determine the potential by integrating. Inverting this process, if we know the potential, we can find the field by differentiating:

This is a vector differential equation; here it is in component form:

55555555555555

E

I 23-7 E Determined from V

55555555555555

E

Example 23-11: for ring and disk.

Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge and (b) a uniformly charged disk.

55555555555555E

I23-8 Electrostatic Potential Energy; the Electron Volt

The potential energy of a charge in an electric potential is U = qV. To find the electric potential energy of two charges, imagine bringing each in from infinitely far away. The first one takes no work, as there is no field. To bring in the second one, we must do work due to the field of the first one; this means the potential energy of the pair is:

I

One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt:

1 eV = 1.6 × 10-19 J.

The electron volt is often a much more convenient unit than the joule for measuring the energy of individual particles.

23-8 Electrostatic Potential Energy; the Electron Volt

I

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

24-1 Capacitors

I

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

24-1 Capacitors

I

When a capacitor is connected to a battery, the charge on its plates is proportional to the voltage:

The quantity C is called the capacitance.

Unit of capacitance: the farad (F):

1 F = 1 C/V.

24-1 Capacitors

I24-2 Determination of Capacitance

For a parallel-plate capacitor as shown, the field between the plates is

E = Q/ε0A.

Integrating along a path between the plates gives the potential difference:

Vba = Qd/ε0A.

This gives the capacitance:

I 24-2 Determination of Capacitance

Example 24-1: Capacitor calculations.

(a) Calculate the capacitance of a parallel-plate capacitor whose plates are 20 cm × 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, given the same air gap d.