Electric Potential Energy and Potential Definitions Examples A B q E 235Unucleus + n (92 protons) Ba Kr (56 p) (36 p) d = 14.6 •10-15m

Conservation of Energy of a particle Kinetic Energy (K) non-relativistic Potential Energy (U) determined by force law for Conservative Forces: K+U is constant total energy is always constant examples of conservative forces gravity; gravitational potential energy springs; coiled spring energy (Hooke’s Law): U(x)=kx2 electric; electric potential energy (today!) examples of non-conservative forces (heat) friction viscous damping (terminal velocity)

Electric potential energy Imagine two positive charges, one with charge Q1, the other with charge Q2: Initially the charges are very far apart, so we say that the initial energy Ui of interaction is zero (we are free to define the energy zero somewhere). If we want to push the particles together, this will require work (since they want to repel).  the final energy Uf of the system will increase by the same amount: DU = Uf – Ui = DW Q1 Q2 Q1 Q2

Electric potential energy Pretend Q1 is fixed at the origin. What is the work required to move Q2, initially at infinity, a distance r away? Q1 r Q2 Remember – work is force times distance: What if Q2 were negative (but Q1 still positive)? Then the work “required” by us would be negative  the charges would like to come together. In this case the final energy is negative! Particles will move to minimize the final potential energy.

Electric Potential Energy In addition to discussing the energy of a “test” charge in a Coulomb field, we can speak of the electric potential energy of the field itself! Reasons? Work had to be done to assemble the charges (from infinity) into their final positions. This work is the potential energy of the field. The potential energy of a system of N charges is defined to be the algebraic sum of the potential energy for every pair of charges. This theme continues in the course with E in capacitors and B in inductors – these devices store electric and magnetic energy. We will start with a couple of example calculations…

Electric Potential Energy Example 1: Nuclear Fission Ba Kr (56 p) (36 p) d = 14.6 •10-15m (The electrons are all “far” away) 235Unucleus + n (92 protons) What is the potential energy of the two nuclei? = 200 MeV Compare this to the typical energy released in a chemical reaction, ~10eV. By allowing the two fragments to fly apart, this potential energy  kinetic energy  heat  drives a turbine  generate electricity. (What holds the protons together in the nucleus to begin with? Gluons! The “strong force”  very short range, very very strong!)

Electric Potential Energy -q +2q d Example 2: What is the potential energy of this collection of charges? Step 1: Bring in +2q from infinity. This costs nothing. Step 2: Bring in one -q charge. The force is attractive! The work required is negative: Step 3: Bring in 2nd -q charge. It is attracted to the +2q, but repelled from the other -q charge. The total work (all 3 charges) is A negative amount of work was required to bring these charges from infinity to where they are now (i.e., the attractive forces between the charges are larger than the repulsive ones).

Consider the 3 collections of point charges shown below. Which collection has the smallest potential energy? (a) (b) (c) d -Q +Q

Consider the 3 collections of point charges shown below. Which collection has the smallest potential energy? (a) (b) (c) d -Q +Q We have to do positive work to assemble the charges in (a) since they all have the same charge and will naturally repel each other. In (b) and (c), it’s not clear whether we have to do positive or negative work since there are 2 attractive pairs and one repulsive pair. (a) (b) (c) U

Preflight 5: A Two charges which are equal in magnitude, but opposite in sign are placed at equal distances from point A. 2) If a third charge is added to the system and placed at point A, how does the electric potential energy of the charge collection change ? a) increases b) decreases c) doesn’t change

Electric potential V(x,y,z) = U(x,y,z)/q0 (U = qV) Consider that we have three charges fixed in space. Q1 Q2 Q3 The potential energy of an added test charge q0 at point P is just q0 r1p r2p r3p Note that this factors: q0 x (the effects of all other charges) Just as we previously defined the electric field as the force/charge, we now define the electric potential as the potential energy/charge: V(x,y,z) = U(x,y,z)/q0 (U = qV) U depends on what qo is, but V is independent of qo (can be + or -) Units of electric potential are volts: 1 V = 1 J/C V(x,y,z) is a scalar field, defined everywhere in space.

Electric potential difference, in terms of E Suppose charge q0 is moved from pt A to pt B through a region of space described by electric field E. A B q E Felec Fwe supply = -Felec Force on the charge due to E  work WAB≡WAB will have to be done to accomplish this task: Þ To get a positive test charge from lower potential to higher potential you need to invest energy - you need to do work The overall sign of this: A positive charge would “fall” from a higher potential to a lower one If a positive charge moves from high to low potential, it can do work on you; you do “negative work” on the charge

Preflight 5: E A B C 4) Points A, B, and C lie in a uniform electric field. What is the potential difference between points A and B? ΔVAB = VB - VA a) ΔVAB > 0 b) ΔVAB = 0 c) ΔVAB < 0 5) Point C is at a higher potential than point A. True False

Preflight 5: 9) A positive charge is released from rest in a region of electric field. The charge moves: a) towards a region of smaller electric potential b) along a path of constant electric potential c) towards a region of greater electric potential

13) If you want to move in a region of electric field without Preflight 5: 13) If you want to move in a region of electric field without changing your electric potential energy. How would you choose the direction ? You would have to move perpendicular to the field if you wish to move without changing electric potential.

A single charge ( Q = -1mC) is fixed at the origin A single charge ( Q = -1mC) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m. What is the sign of the potential difference between A and B? (Δ VAB º VB - VA ) x -1mC ´ A B (a) ΔVAB < 0 (b) Δ VAB = 0 (c) Δ VAB > 0

You could also determine the sign directly from the definition: A single charge ( Q = -1mC) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m. What is the sign of the potential difference between A and B? (Δ VAB º VB - VA ) x -1mC ´ A B (a) ΔVAB < 0 (b) Δ VAB = 0 (c) Δ VAB > 0 The simplest way to get the sign of the potential difference is to imagine placing a positive charge at point A and determining which way it would move. Remember that a positive charge will always “fall” to lower potential. A positive charge at A would be attracted to the -1mC charge; therefore NEGATIVE work would be done to move the charge from A to B. Since , ΔVAB <0 !! You could also determine the sign directly from the definition:

Δ VAB is Independent of Path q E The integral is the sum of the tangential (to the path) component of the electric field along a path from A to B. This integral does not depend upon the exact path chosen to move from A to B. Δ VAB is the same for any path chosen to move from A to B (because electric forces are conservative).

Does it really work? A C B E h r q Consider case of constant field: Direct: A - B Long way round: A - C – B A C B E h r q dl So here we have at least one example of a case in which the integral is the same for BOTH paths. In fact, it works for all paths  see Appendix A.

Preflight 5: E A B C 7) Compare the potential differences between points A to C and points B to C. a) VAC > VBC b) VAC = VBC c) VAC < VBC

3 (a) WAB < 0 (c) WAB > 0 (b) WAB = 0 A positive charge Q is moved from A to B along the path shown by the arrow. What is the sign of the work done to move the charge from A to B? A B (a) WAB < 0 (b) WAB = 0 (c) WAB > 0

3 (a) WAB < 0 (c) WAB > 0 (b) WAB = 0 A positive charge Q is moved from A to B along the path shown by the arrow. What is the sign of the work done to move the charge from A to B? A B (a) WAB < 0 (b) WAB = 0 (c) WAB > 0 A direct calculation of the work done to move a positive charge from point A to point B is not easy. Neither the magnitude nor the direction of the field is constant along the straight line from A to B. But, you DO NOT have to do the direct calculation. Remember: potential difference is INDEPENDENT OF THE PATH!! Therefore we can take any path we wish. Choose a path along the arc of a circle centered at the charge. Along this path at every point!!

Electric Potential: where is it zero? So far we have only considered potential differences. Define the electric potential of a point in space as the potential difference between that point and a reference point. a good reference point is infinity ... we often set V = 0 the electric potential is then defined as: for a point charge at origin, integrate in from infinity along some axis, e.g., the x-axis here “r” is distance to origin line integral Potential from a point charge q

Potential from N charges The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately (this is much easier to calculate than the net electric field, which would be a vector sum). Q1 Q2 Q3 P r1p r2p r3p Þ 4 Appendix B: Calculation of potential from a dipole

Preflight 5: +5 μC -3 μC A 11) Two charges q1 = + 5 μC, q2 = -3μC are placed at equal distances from the point A. What is the electric potential at point A? a) VA < 0 b) VA = 0 c) VA > 0

Which of the following charge distributions produces V(x) = 0 for all points on the x-axis? (we define V(x) º 0 at x = ¥ ) +2mC +1mC +2mC +1mC +2mC -2mC x x x -2mC -1mC -2mC -1mC +1mC -1mC (c) (a) (b)

Which of the following charge distributions produces V(x) = 0 for all points on the x-axis? (we define V(x) º 0 at x = ¥ ) +2mC +1mC +2mC +1mC +2mC -2mC x x x -2mC -1mC -2mC -1mC +1mC -1mC (c) (a) (b) The key here is to realize that to calculate the total potential at a point, we must only make an ALGEBRAIC sum of the individual contributions. Therefore, to make V(x)=0 for all x, we must have the +Q and -Q contributions cancel, which means that any point on the x-axis must be equidistant from +2mC and -2mC and also from +1mC and -1mC. This condition is met only in case (a)!

Summary Potential energy stored in a static charge distribution work we do to assemble the charges Electric potential energy of a charge in the presence of a set of source charges potential energy of the test charge equals the potential from the sources times the test charge: U = qV If we know the electric field E, allows us to calculate the potential function V everywhere (define VA = 0 above) Potential due to n charges:

Appendix A: Independent of Path? B Consider any path from A to B as being made up of a succession of arc plus radial parts as above. The work along the arcs will always be 0 (since ), leaving just the sum of the radial parts, e.g.: In going from rA to rB, all terms with intermediate radii will cancel, leaving just the initial and final radii: Therefore it's general! r2 A r1 q

Appendix B: Electric Dipole z a q +q -q r 1 2 The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms. r2-r1 Rewrite this for special case r>>a: Þ In Lecture 7 you will see how to use this potential to calculate the E field of a dipole (remember how messy the direct calculation was?)