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.

Referred by the symbol V 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 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 HEADLIGHT The 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 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 ENERGY A particle has a mass of 1 EXAMPLE 4: THE CONSERVATION OF ENERGY A 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?

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Section 3 and 4 notes Take your own notes. I will check at the end of class.

19.3 THE ELECTRIC POTENTIAL DIFFERENCE CREATED BY POINT CHARGES When the test charge moves from A to B, work is done by this force. R varies between ra and rb, the force F also varies, and the work is not the product of the force and the distance between the points. Work can be found with the methods of integral calculus.

This equation can be used whether q is positive is negative, and whether qo is positive or negative. W can be substituted with the change in V.

Point B is located farther and farther from the charge q, rb becomes larger and larger. Limit that rb is infinitely large, the term kq/rb becomes zero, and it is customary to set Vb equal to zero also. In this limit eq. 19.5 becomes Va = kq/ra, and subscripts can be omitted and the following equation can be used. Potential of a point charge.

V does not refer to the potential, rather V = kq/r stands for the amount by which the potential at a distance r from a point charge differs from the potential at an infinite distance away. V refers to a potential difference with the assumption that the potential at infinity is zero. When q is positive so is V and same w/ negative q.

EXAMPLE 5: THE POTENTIAL OF A POINT CHARGE Using a zero reference potential at infinity, determine the amount by which a point charge of 4.0 x 10^-8 C alters the electric potential at a spot 1.2 m away when the charge is (a) positive and (b) negative.

A single point charge raises or lowers the potential at a given location, depending on whether the charge is positive or negative. When two or more charges are present, the potential due to all the charges is obtained by adding together the individual potentials.

EXAMPLE 6: THE TOTAL ELECTRIC POTENTIAL At locations A and B, find the total electric potential due to the two point charges.

EXAMPLE 8: THE POTENTIAL ENERGY OF A GROUP OF CHARGES Three point charges; initially, they are infinitely far apart. They are then brought together and placed at the corners of an equilateral triangle. Each side of the triangle has a length of 0.50m. Determine the electric potential energy of the triangular group. In other words, determine the amount by which the electric potential energy of the group differs from that of the three charges in their initial, infinitely separated locations.

19.4 EQUIPOTENTIAL SURFACES AND THEIR RELATION TO THE ELECTRIC FIELD Equipotential surface: surface on which the electric potential is the same everywhere. Equipotential surface around an isolated point charge. Surface is a spherical surface. The larger the distance r, the smaller is the potential of the equipotential surface.

The net electric force does no work as a charge moves on an equipotential surface. Potential remains the same Va = Vb, therefore W will be zero. (Vb – Va = -Wab/qo) Euipotential sphere: the electric field is perpendicular to the surface. Points outward in the direction of decreasing potential.

ELECTRIC DIPOLE

EXAMPLE 9: THE ELECTRIC FIELD AND POTENTIAL ARE RELATED The plates of the capacitor are separated by a distance of 0.032m, and the potential difference between them is -64V. Between the two labeled equipotential surfaces there is a potential difference of -3.0V. Find the spacing between the two labeled surfaces.

Parallel plate capacitor. A higher potential B lower potential V = Va – Vb D = separation, displacement from plate A to plate B. D = sb - sa

11. Two charges A and B are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge A is 0.18m from the spot, while charge B is 0.43m from it. Find the ratio qb/qa of the charges.

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.75mm. When an electric spark jumps between them, the magnitude of the electric field is 4.7 x 10^7 V/m. What is the magnitude of the potential difference between the conductors? 27.

Homework Pg. 599-600 12,13,28,29 Quiz on sections 1-4 on Wednesday