Chapter 18 Electric Potential Chapter 23 opener. We are used to voltage in our lives—a 12-volt car battery, 110 V or 220 V at home, 1.5 volt flashlight batteries, and so on. Here we see a Van de Graaff generator, whose voltage may reach 50,000 V or more. Voltage is the same as electric potential difference between two points. Electric potential is defined as the potential energy per unit charge. The children here, whose hair stands on end because each hair has received the same sign of charge, are not harmed by the voltage because the Van de Graaff cannot provide much current before the voltage drops. (It is current through the body that is harmful, as we will see later.)

Electric Potential (V) Point Charges, Continuous Charge Distributions Chapter Outline Electric Potential Energy Potential Difference Relation between Electric Potential (V) & Electric Field (E) Electric Potential due to: Point Charges, Continuous Charge Distributions

Potential due to Charge Distributions Equipotential Surfaces, Electric Dipole Potential E Determined from V Electrostatic Potential Energy: Applications: Cathode Ray Tubes, TV & Computer Monitors, Oscilloscopes

Brief Review of Some Physics I Concepts Definition: A force is conservative if & only if: The work done by that force on an object moving from one point to another depends ONLY on the initial & final positions of the object, & is independent of the particular path taken. Example Gravity Figure 8-1. Caption: Object of mass m: (a) falls a height h vertically; (b) is raised along an arbitrary two-dimensional path.

Conservative Force: Another definition! A force is conservative if & only if the net work done by the force on an object moving around any closed path is zero. Figure 8-2. Caption: (a) A tiny object moves between points 1 and 2 via two different paths, A and B. (b) The object makes a round trip, via path A from point 1 to point 2 and via path B back to point 1.

Friction is a Nonconservative Force! Friction Force: If friction is present, work done depends on the starting & ending points, AND on the path taken. Friction is a Nonconservative Force! Figure 8-3. Caption: A crate is pushed at constant speed across a rough floor from position 1 to position 2 via two paths, one straight and one curved. The pushing force FP is always in the direction of motion. (The friction force opposes the motion.) Hence for a constant magnitude pushing force, the work it does is W = FPd, so if d is greater (as for the curved path), then W is greater. The work done does not depend only on points 1 and 2; it also depends on the path taken. Friction Force: The work done by friction depends on the path!

U  Energy associated with the position or configuration of a mass. Potential Energy A mass can have a Potential Energy due to its environment. Potential Energy  U  Energy associated with the position or configuration of a mass. NOTE! In our book, the symbol for Potential Energy is PE. So, sometimes, in what follows, I’ll use U  PE

Systems with Potential Energy: U (or PE)  Energy associated with a position or configuration of a mass. Examples: Systems with Potential Energy: 1. A wound-up spring 2. A stretched elastic band 3. An object at some height above the ground

Can only be defined for Conservative Forces! Potential Energy: Can only be defined for Conservative Forces! Another Physics I Result: The change in the Potential Energy is defined to be the negative of the work done by the conservative force. End of Brief Review!!

Electric Potential Energy Consider a point charge in an electric field E. We know that it experiences a Coulomb force: F = qE For simplicity, assume that E is along the x axis. Suppose that, under the influence of this force, the charge moves a distance Δx The work done by the electric force on the charge is W = F Δx = qE Δx

W = F Δx = qE Δx The work done by the electric force on the charge is The electric force is conservative, so the work done is independent of the path & a potential energy exists. If the electric force does an amount of work W on the charged particle, there is an accompanying change in electric potential energy.

ΔPEelec = -W = - FΔx ΔPEelec = -qEΔx The electric potential energy is denoted as PEelec The change in PEelec for the point charge is: ΔPEelec = -W = - FΔx ΔPEelec = -qEΔx Because F is conservative, the change in potential energy depends on the endpoints of the motion, but not on the path taken.

From Physics I, we know that potential energy can be viewed as stored energy in the system. A positive energy ΔPEelec can be stored in a system that is composed of the charge & the field. Stored energy can be taken out of the system. This energy may show up as an increase in the kinetic energy, ΔKE of the particle

Potential Energy: Two Point Charges From Coulomb’s Law, q1 & q2 exert forces on each other: If they are like charges, they will repel each other. If they are unlike charges, they will attract each other.

The electric potential energy is The electric force is The electric potential energy is Note that PEelec varies as 1/r while the force varies as 1/r2

Note that PEelec approaches zero when the two charges are very far apart. That is, when r becomes infinitely large. The force also approaches zero in this limit. As in Physics I, only changes in potential energy are important. For two point charges, this has the form:

PEelec & Superposition The results for 2 point charges can be extended by using the superposition principle. If there is a collection of point charges, the total potential energy is the sum of the potential energies of each pair of charges.

PEelec & Superposition Complicated charge distributions can always be treated as a collection of point charges arranged in some particular manner. The electric forces between a collection of charges will always be conservative.

Point Charge in a Constant E Field For a point charge in a constant electric field E. We know that the Coulomb force is: F = qE Assume that E is along the x axis. Under the influence of this force, the charge moves a distance Δx

Point Charge in a Constant E Field The work done by the electric force on the charge when it moves Δx is W = F Δx = qE Δx If the electric force does an amount of work W on the charged particle, there is an accompanying change in electric potential energy.

Electrostatic Potential Energy, Potential Difference The Electrostatic Force is Conservative, so an Electrostatic Potential Energy can be defined. As in Physics I, the change in electric potential energy is negative of the work done by the electric force: Figure 23-1. Work is done by the electric field in moving the positive charge q from position a to position b.

The Electric Potential V is defined as the potential energy per unit charge: The SI Unit of Electric Potential is The Volt (V). 1 V = 1 J/C. Only changes in potential can be measured, So we can arbitrarily choose the point where V = 0.

1 eV  1.60  10-19 J Electron-Volts Electron-Volt (eV) PE = 1 V. A unit of energy that is commonly used in atomic & nuclear physics is the Electron-Volt (eV) One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (electron or proton) is moved through a potential difference PE = 1 V. 1 eV  1.60  10-19 J

Potential Difference in a Uniform E Field The relation for the electric potential difference between two points A and B can be simplified If the electric field E is uniform (constant): The displacement vector d points from A to B & is parallel to the field lines. The negative sign says that the electric potential at point B is lower than at point A. Electric field lines always point in the direction of decreasing electric potential.

Potential Difference in a Uniform E Field When E is directed down, point B is at a lower potential than A. When a positive test charge moves from A to B, the charge-field system loses potential energy. Electric field lines always point in the direction of decreasing electric potential.

Conceptual Example: A Negative Charge. Suppose that a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change? Figure 23-2. Central part of Fig. 23–1, showing a negative point charge near the negative plate, where its potential energy (PE) is high. Example 23–1. Solution: The electron will move towards the positive plate if released, thereby increasing its kinetic energy. Its potential energy must therefore decrease. However, it is moving to a region of higher potential V; the potential is determined only by the existing charge distribution and not by the point charge. U and V have different signs due to the negative charge.

height have gravitational Analogy between Gravitational Potential Energy & Electrical Potential Energy 2 masses m & 2m, at the same height have gravitational potential energies ; U1 = mgh, U2 = 2mgh. Clearly, U2 = 2U1 > U1 Figure 23-3. (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.

Analogy between Gravitational Potential Energy & Electrical Potential Energy 2 masses m & 2m, at the same height have gravitational potential energies; U1 = mgh, U2 = 2mgh. Clearly, U2 = 2U1 > U1 2 charges Q & 2Q, are at the same electric potential Vba but they have different electric potential energies U1 = QVba, U2 = 2QVba. Clearly, U2 = 2U1 > U1 Figure 23-3. (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.

Electrical Sources Like batteries & generators supply a constant potential difference. On the right are some typical potential differences, both natural & manufactured:

Example Electron in a CRT Vb – Va = Vba = +5,000 V An electron in a cathode ray tube (CRT) is accelerated from rest through a potential difference Vb – Va = Vba = +5,000 V Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

Example Electron in a CRT Vb – Va = Vba = +5,000 V Calculate: An electron in a cathode ray tube (CRT) is accelerated from rest through a potential difference Vb – Va = Vba = +5,000 V Calculate: (a) The change in electric potential energy of the electron. Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

U = qVba = (-1.6  10-19)(5000) = -8.0  10-16 J Example Electron in a CRT An electron in a cathode ray tube (CRT) is accelerated from rest through a potential difference Vb – Va = Vba = +5,000 V Calculate: (a) The change in electric potential energy of the electron. U = qVba = (-1.6  10-19)(5000) = -8.0  10-16 J Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

U = qVba = (-1.6  10-19)(5000) = -8.0  10-16 J Example Electron in a CRT An electron in a cathode ray tube (CRT) is accelerated from rest through a potential difference Vb – Va = Vba = +5,000 V Calculate: (a) The change in electric potential energy of the electron. U = qVba = (-1.6  10-19)(5000) = -8.0  10-16 J (b) The speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration. Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

U = qVba = (-1.6  10-19)(5000) = -8.0  10-16 J Example Electron in a CRT An electron in a cathode ray tube (CRT) is accelerated from rest through a potential difference Vb – Va = Vba = +5,000 V Calculate: (a) The change in electric potential energy of the electron. U = qVba = (-1.6  10-19)(5000) = -8.0  10-16 J (b) The speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration. Conservation of Mechanical Energy: K + U = 0 or K = -U. So (½)mv2 – 0 = - qVba = - 8.0  10-16 J so, v = 4.2  107 m/s Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.

Solution: a. The change in potential energy is qV = -8.0 x 10-16 J. b. The change in potential energy is equal to the change in kinetic energy; solving for the final speed gives v = 4.2 x 107 m/s.