Physics 102: Lecture 3 Electric Potential

Recall from last lecture…. E field has magnitude and direction: E due to point charge Q: E=kQ/r2 Force on charge q due to E: F=qE E and F are vectors Electric field lines Density gives strength (# proportional to charge.) Arrow gives direction (Start + end on -) Conductors E=0 inside a conductor

Overview for Today’s Lecture Electric Potential Energy/ Work Uniform fields Point charges Electric Potential (like height) Show large and small battery 9 volt small, 6 volt large 07

Recall Work from P101 Work done by the force given by: W = F d cos(q) Positive: Force is in direction moved Negative: Force is opposite direction moved Zero: Force is perpendicular to direction moved Careful ask WHAT is doing work! Opposite sign for work done by you! Conservative Forces (not friction) Change in Potential Energy = -Wconservative Use book or brick for prop. Ask students if I am doing positive or negative work, what about gravity. 09

Work and D Potential Energy Gravity Uniform Electric Field Brick raised yi yf +charge q moved d to left FE = qE (right) WE = -qEd DUE= +qEd FG = mg (down) WG = -mgh DUG= +mgh yf Fg=mg Fg=mg q d Fg=mg Uniform E q h Fg=mg Fg=mg Fg=mg Fg=mg yi Fg=mg 20

Work and D Potential Energy W = F d cos(q) Gravity Electric Brick raised yi yf + Charge moved ∞  rf FE = kq1q2/r2 WE = -kq1q2/rf DUE= +kq1q2/rf FG = mg (down) WG = -mgh DUG= +mgh yf rf h yi 20

Preflight 3.1 ACT Uniform E B - Uniform E In what direction does the force on a negative charge at point A point? left right up 53% 43% 4% Force on charge is in same direction of field if POSITIVE and opposite direction if NEGATIVE. 10

Preflight 3.2 motion - F C - F “I would say zero because the path is perpendicular to the field” - F - F - F A B Uniform E When a negative charge is moved from A to C the ELECTRIC force does positive work. zero work. negative work. 10% 85% 05% 11

Preflight 3.3 ACT Uniform E B “because the direction of the displacement is 180 degrees from direction of the force ” - F - F - F - F - F - Uniform E motion When a negative charge is moved from A to B the ELECTRIC force does positive work. zero work. negative work. 66% 7% 21% 13

Preflight 3.5 ACT Uniform E B - - - - - Uniform E When a negative charge is moved from A to B, the electric potential energy of the charge Increases is constant decreases D(EPE) = -WE field Electric force did negative work so electric potential energy increased. Just like pushing mass uphill. 33% 14% 53% 14

ACT: Electric Potential Energy AC: W=0 E + C CB: W<0 B A - - - - - When a negative charge is moved from A to B, the electric potential energy of the charge (A) increases (B) is constant (C) decreases Now switch to point charges! 1) The electric force is directed to bring the electron closer to be proton. 2) Since the electron ends up further from the proton the electric field did negative work. 3) So the electric potential energy increased 17

Work done by YOU to assemble 3 charges Example W1 = 0 W2 = k q1 q2 /r =3.6 mJ =(9109)(110-6)(210-6)/5 W3 = k q1 q3/r + k q2 q3/r (9109)(110-6)(310-6)/5 + (9109)(210-6)(310-6)/5 =16.2 mJ Wtotal = +19.8 mJ WE = -19.8 mJ DUE = +19.8 mJ Do this live with balls on the table! 3 5 m 5 m 1 2 5 m Note the units and watch signs: This is like moving mass uphill 24

ACT: Work done by YOU to assemble 3 negative charges How much work would it take YOU to assemble 3 negative charges? Likes repel, so YOU will still do positive work! 3 W = +19.8 mJ W = 0 mJ W = -19.8 mJ 5 m 5 m 1 2 5 m 27

Preflight 3.11 1 + 5 m 5 m Negative because i really hate physics. And i dont know what is being ask in this question. + - 2 5 m 3 The total work required by you to assemble this set of charges is: (1) positive (2) zero (3) negative Bring in (3): zero work Bring in (2): negative work Bring in (1): zero work (see next page for explanation) 57% 16% 27%

Preflight 3.11 + - 5 m 1 3 2 Bring in (3): zero work because the other charges are far away so the electric field due to those charges is zero. Bring in (2): negative work. why? Let’s figure out the work done by the electric field, which is just the negative of the work done. The electric field felt by charge 2 is the field due to charge 3, which points toward charge 3. So, we are moving charge 2 in the same direction of the field. Therefore the work done by the field is positive, so the work done by you is negative Bring in (1): zero work. why? We must do negative work due to charge 3 and an equal amount of positive work due to charge 2, so the net work is zero. Another way to think about it is that the electric potential energy of charge 1 is zero, since it has equal but opposite contributions from charges 2 and3 Net result is the sum of 0,negative, and 0 and is therefore negative.

Preflight 3.11 - + 1 3 2 5 m Yet another way to work the problem: Wyou = -WE = Electric Potential Energy (EPE) of the three charges. EPE = kq1q2/r + kq2q3/r +kq1q3/r where r is the separation between the charges (5 m). All three terms have the same magnitude, since all the charges have the same magnitude. The first term is positive but the next two are negative. Therefore, the EPE is negative, so that Wyou is negative. As a practice exercise, try calculating Wyou, assuming the magnitude of each charge is 5 C. Answer = -0.045 J.

Work and D Electric Potential Gravity Uniform Electric Field Brick raised yi yf +charge q moved d to left DUE= +qEd = change in electric potential energy DUE= +q(Ed) = charge(q) x change in electric potential (Ed) VE Moving opposite to E increases VE DUG= +mgh = change in gravitational potential energy DUG = m(gh) = mass (m) x change in gravitational potential (gh) VG Moving from low to high increases VG yi yf h m Uniform E q d m 20

Electric Potential Units Joules/Coulomb Volts (symbol V) Batteries Outlets EKG Really Potential differences Equipotential lines at same “height” Field lines point “downhill” from higher to lower potential V = k q/r (distance r from charge q) V(∞) = 0 Comment on lab w/ equipotential lines 31

Preflight 3.7 Electric field points from greater potential to lower potential The electric potential at point A is _______ at point B greater than equal to less than 55% 26% 19% 32

Preflight 3.9 89% conductor The electric potential at point A is _______ at point B greater than equal to less than “The electric field within a conductor is zero, and therefore, the potential for points A and B are the same 33

Preflight Summary Path Vfinal - Vinitial Charge WE field D U = q DV + Negative A  B Positive Negative Positive Negative + - A  C Zero C  B Negative + - 35

ACT: Electric Potential + C B A The electric potential at A is ___________ the electric potential at B. greater than equal to less than 1) Electric field lines point “down hill” 2) AC is equipotential path (perpendicular to E) 3) CB is down hill, so B is at a lower potential than (“down hill from”) A 38

Electric Potential: Summary E field lines point from higher to lower potential For positive charges, going from higher to lower potential is “downhill” For negative charges, going from lower to higher potential is “downhill” For a battery, the + terminal is at a higher potential than the – terminal Positive charges tend to go “downhill”, from + to - Negative charges go in the opposite direction, from - to +

Comparison: Electric Potential Energy vs. Electric Potential Electric Potential Energy (U) - the energy of a charge at some location. Electric Potential (V) - found for a location only – tells what U would be if a charge were located there: U = Vq Usually we talk only about changes in potential or potential energy when moving from one location to another Neither U nor V has direction, just location. Sign matters!

Electric Potential due to a point charge Example not covered during lecture Electric Potential due to a point charge What is the electric potential V a distance r from a point charge, assuming V=0 at  answer: V = k q/ r 42

Two Charges Example not covered during lecture Calculate electric potential at point A due to charges Calculate V from +7mC charge Calculate V from –3.5mC charge Add (EASY!) V = kq/r V7=(9109)(710-6)/5 = 12.6103V V3=(9109)(-3.510-6)/5 = -6.3103V Vtotal = V7+V3 = +6.3103V A 4 m 6 m Q=+7.0mC Q=-3.5 mC W=DU=DVq =(+6.3103V)(2mC) =+12.6 mJ How much work do you have to do to bring a 2 mC charge from far away to point A? 46

not covered during lecture ACT: Two Charges In the region II (between the two charges) the electric potential is 1) always positive 2) positive at some points, negative at others. 3) always negative I II III Q=+7.0mC Q=-3.5 mC Very close to positive charge potential is positive Very close to negative charge potential is negative 48

To Do Read 17.5-6 Extra problems from textbook Ch 17: Concepts 1-8 Problems 1, 9, 15, 19, 23, 25 Bring “Problem Solver” to discussion section Complete preflight before Monday 6:00am. 50