ELECTRIC POTENTIAL ENERGY AND ELECTRIC POTENTIAL

Pg. 573 solar car• An electric potential energy exists that is

comparable to the gravitational potential energy.

• Review of gravitational potential energy.• Fig. 19.1• The work Wab done by the gravitational force

when the ball falls from a height of ha to a height of hb is.

• Work done by the gravitational force equals the initial gravitational potential energy minus the final gravitational potential energy.

• Fig. 19.2• Test charge experiences an electric force that

is directed downward.• As the charge moves from A to B, work is done

by this force, same way work is done by the gravitational force.

19.2 THE ELECTRIC POTENTIAL DIFFERENCE• F = qoE• The work that it does as the charge moves

from A to B depends on the charge qo.• Can be expressed per-unit-charge basis, by

dividing both sides by the charge.

• EPE/qo is the electric potential energy per unit charge ----- called the electric potential or the potential

• Referred by the symbol V

The electric potential V at a given point is the electric potential energy EPE of a small test charge qo situated at the point divided by the charge itself:

• SI unit of electric potential is a joule per coulomb, known as a volt.

• Alessandro Volta, invented voltaic pile (battery).

• EPE and volt are not the same.• Electric potential energy is an energy and is

measured in joules.• Electric potential is an energy per unit charge

and is measured in joules per coulomb, or volts.

• We can now relate the work Wab done by the electric force when a charge qo moves from A to B to the potential difference Vb-Va between the points. Combining equations 19.2 & 19.3

EXAMPLE 1: WORK, ELECTRIC POTENTIAL ENERGY, AND ELECTRIC POTENTIAL

the work done by the electric force as the test charge (qo = +2.0 x 10^-6C) moves from A to B is Wab = +5.0 x 10^-5J. (a) Find the value of the difference, ΔEPE = EPEb – EPEb, in the electric potential energies of the charge between these points. (b) Determine the potential difference, ΔV = Vb – Va, between the points.

• The positive charge in fig. 19.2 accelerates as it moves from A to B because of the electric repulsion from the upper plate and the attraction to the lower plate. Since point A has a higher electric potential than point B, we conclude that a positive charge accelerates from a region of higher electric potential toward a region of lower electric potential

• A negative charge placed between the plates behave in the opposite fashion.

• A negative charge accelerates from a region of lower potential toward a region of higher potential.

• Car Battery• + (a)higher potential than – (b) Va – Vb = 12V

EXAMPLE 3: OPERATING A HEADLIGHTThe wattage of the headlight on the last slide is

60W. Determine the number of particles, each carrying a charge of 1.60 x 10^-19 C (the magnitude of the charge on an electron), that pass between the terminals of the 12 V car battery when the headlight burns for one hour.

• The number of particles is the total charge that passes between the battery terminals in one hour divided by the magnitude of the charge on each particle. The total charge is that needed to convey the energy used by the headlight in one hour. This energy is related to the wattage of the headlight, which specifies the power or rate at which energy is used, and the time the light is on.

WHAT YOU KNOW (about this problem)

Volt is used for measuring electric potential difference.

Volt can be used to describe an electron or a proton, electron volt (eV).

One electron volt is the magnitude of the amount by which the potential energy of an electron changes when the electron moves through a potential difference of one volt.

• One million (10^+6) electrons volts of energy is referred to as one MeV, and one billion (10^+9) electron volts of energy is one GeV, where the G stands for the prefix “giga.”

EXAMPLE 4: THE CONSERVATION OF ENERGYA particle has a mass of 1.8 x 10^-5 kg and a charge

of +3.0 x 10^-5C. It is released from rest at point A and accelerates until it reaches point B. The particle moves on a horizontal straight line and does not rotate. The only forces acting on the particle are the gravitational force and an electrostatic force. The electric potential at A is 25V greater than that at B; in other words Va – Vb = 25V.

What is the translational speed of the particle at point B?