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LECTURE 4 POTENTIAL September 20, 2003 Alternate Lecture Titles Back to Physics 2048 You can run but you can’t hide! The PHY 2048 Brain Partition To move the mass m from the ground to a point a distance h above the ground requires that work be done on the particle. h h B m Reference “0” W mgdy mgh 0 W is the work done by an external force. mgh represents this amount of work and is the POTENTIAL ENERGY of the mass at position h above the ground. A The reference level, in this case, was chosen as the ground but since we only deal with differences between Potential Energy Values, we could have chosen another reference. Let’s Recall Some more PHY2048 A mass is dropped from a height h above the ground. What is it’s velocity when it strikes the ground? We use conservation of energy to compute the answer. h B m A 1 2 mgh (0) (0) mv 2 and v 2 gh Result is independent of the mass m. Using a different reference. y y=h E PE KE B m y=b (reference level) y=0 A Still falls to here. 1 2 mg (h b) 0 mg (b) mv 2 1 2 mgh mgb mgb mv 2 v 2 gh Energy Methods Often easier to apply than to solve directly Newton’s law equations. Only works for conservative forces. One has to be careful with SIGNS. VERY CAREFUL! I need some help. THINK ABOUT THIS!!! When an object is moved from one point to another in an Electric Field, It takes energy (work) to move it. This work can be done by an external force (you). You can also think of this as the FIELD doing the negative of this amount of work on the particle. Let’s look at it: move a mass from yi to yf Change in potential energy due to external force: yf W force dist . W mg ( y f yi ) ( PE ) Negative of the work done BY THE FIELD. External Field yi W (mg ) ( y f yi ) ( PE ) Keep it! And also remember: The net work done by a conservative (field) force on a particle moving around a closed path is ZERO! A nice landscape Work done by external force = mgh How much work here by gravitational field? h mg The gravitational case: Someone else’s path IMPORTANT The work necessary for an external agent to move a charge from an initial point to a final point is INDEPENDENT OF THE PATH CHOSEN! The Electric Field Is a conservative field. No frictional losses, etc. Is created by charges. When one (external agent) moves a test charge from one point in a field to another, the external agent must do work. This work is equal to the increase in potential energy of the charge. It is also the NEGATIVE of the work done BY THE FIELD in moving the charge from the same points. A few things to remember… A conservative force is NOT a Republican. An External Agent is NOT 007. Electric Potential Energy When an electrostatic force acts between two or more charged particles, we can assign an ELECTRIC POTENTIAL ENERGY U to the system. Example: NOTATION U=PE HIGH U LOWER U E q F A B d Work done by FIELD is Fd Negative of the work done by the FIELD is -Fd Change in Potential Energy is also –Fd. The charge sort-of “fell” to lower potential energy. Gravity Negative of the work done by the FIELD is –mg h = U mg Bottom Line: Things tend to fall down and lower their potential energy. The change, Uf – Ui is NEGATIVE! Electrons have those *&#^ negative signs. Electrons sometimes seem to be more difficult to deal with because of their negative charge. They “seem” to go from low potential energy to high. They DO! They always fall AGAINST the field! Strange little things. But if YOU were negative, you would be a little strange too! An Important Example Designed to Create Confusion or Understanding … Your Choice! The change in potential energy Final position of the electron is the negative of the work done by the field in moving the electron from the initial position to the final position. FORCE d W F ( y f yi ) E U W (e) ( E ) ( y f yi ) Initial position e A sad and confused Electron. negative! negative charge Force against The direction of E An important point In calculating the change in potential energy, we do not allow the charge to gain any kinetic energy. We do this by holding it back. That is why we do EXTERNAL work. When we just release a charge in an electric field, it WILL gain kinetic energy … as you will find out in the problems! AN IMPORTANT DEFINITION Just as the ELECTRIC FIELD was defined as the FORCE per UNIT CHARGE: F E q VECTOR We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE: U V q SCALAR UNITS OF POTENTIAL U Joules V VOLT q Coulomb Watch those #&@% (-) signs!! The electric potential difference V between two points I and f in the electric field is equal to the energy PER UNIT CHARGE between the points: U i U W V V f Vi q q q q Uf Where W is the work done BY THE FIELD in moving the charge from One point to the other. Let’s move a charge from one point to another via an external force. The external force does work on the particle. The ELECTRIC FIELD also does work on the particle. We move the particle from point i to point f. The change in kinetic energy is equal to the work done by the applied forces. K K f K i Wapplied W field if K 0 Wapplied W field also U U f U i Wapplied Furthermore… U Wapplied V q q so Wapplied qV If we move a particle through a potential difference of V, the work from an external “person” necessary to do this is qV Example Electric Field = 2 N/C 1 mC d= 100 meters Work done by EXTERNAL agent Change in potential Energy. PE qEd 1mC 2( N / C ) 100m 2 10 4 Joules One Step More Work done by EXTERNAL agent Change in potential Energy. PE qEd 1mC 2( N / C ) 100m 2 10 4 Joules PE Change in POTENTIAL q 2 104 Joules J V 200 200 Volts 6 110 C C The Equipotential Surface DEFINED BY V 0 It takes NO work to move a charged particle between two points at the same potential. The locus of all possible points that require NO WORK to move the charge to is actually a surface. Example: A Set of Equipotenital Surfaces Back To Yesteryear Field Lines and Equipotentials Electric Field Equipotential Surface Components Enormal Electric Field x Eparallel Work to move a charge a distance x along the equipotential surface Is Q x Eparallel X x Equipotential Surface BUT This an EQUIPOTENTIAL Surface No work is needed since V=0 for such a surface. Consequently Eparallel=0 E must be perpendicular to the equipotential surface Therefore E E E V=constant Field Lines are Perpendicular to the Equipotential Lines Equipotential Work external q0 (V f Vi ) Consider Two Equipotential Surfaces – Close together Work to move a charge q from a to b: dWexternal Fapplied ds qEds also b a E dWexternal q(V dV ) V qdV ds Eds dV V+dV V and dV E ds Vector...E V Where i j k x y z Typical Situation A Brief Review of Concept The creation of a charged particle distribution creates ELECTRICAL POTENTIAL ENERGY = U. If a system changes from an initial state i to a final state f, the electrostatic forces do work W on the system U U f U i W This is the NEGATIVE of the work done by the field. Calculation E q i f • An external force F is necessary to move the charge q from i to f. The work done by this external force is also equal to the change in potential energy of the charged particle. Note the (-) sign is because F and E are in opposite directions. U Fx qEx • Continuous case: U Fexternaldx For Convenience It is often convenient to set up a particular reference potential. For charged particles interacting with each other, we take U=0 when the particles are infinitely apart. Consequently U=(-) of the work done by the field in moving a particle from infinity to the point in question. Keep in Mind W F d Force and Displacement are VECTORS! Potential is a SCALAR. UNITS 1 VOLT = 1 Joule/Coulomb For the electric field, the units of N/C can be converted to: 1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM) Or 1 N/C = 1 V/m So an acceptable unit for the electric field is now Volts/meter. N/C is still correct as well. In Atomic Physics It is useful to define an energy in eV or electron volts. One eV is the additional energy that an proton charge would get if it were accelerated through a potential difference of one volt. 1 eV = e x 1V = (1.6 x 10-19C) x 1(J/C) = 1.6 x 10-19 Joules. Coulomb Stuff Consider a unit charge (+) being brought from infinity to a distance r from a Charge q: q x r 1 q E 40 r 2 To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate. The math…. ()1 dr W Fexternaldx (1)q 2 4 r 0 r and W q V Q( 1) 40 1 q V 40 r dr q r 1 r r 2 40 (1) r For point charges qi V 40 i ri 1 Example: Find potential at P q1 q2 1 1 V (q1 q2 q3 q4 ) 40 r d 1.3m d r 0.919m 2 d P r q3 q4 q1=12nC q2=-24nC q3=31nC q4=17nC Sq=36 x 10-9C V=350 Volts (check the arithmetic!!) Brief Discussion on Math Integration There are many applications of integration in physics. We use it to add things up over an area, a line, or perhaps a volume. We often use density functions. The Examples we will use will be charge densities. Line of Charge dx dq Charge per unit length m or . m coul/meter dq dx q ( x)dx can be constant or a function of position (x). Area Charge Density s dA=dxdy dy dx charge s area dq sdA The Job dA dxdy dA=dxdy dy dx R R A dxdy R R y R x 2 R 2 1 2 2 1 / 2 dy R x (2 xdx) 2 substitute and go crazy ........ Another Way to do dA (note rhyme) dr rdq dq r q dA= dr rdq r drdq Take a look dA rdrdq A dA rdrdq A 2 A 2 0 0 r rdr dq 0 r2 r2 2 dq 2 r 2 2 Still another view dA 2rdr r R dr A 2 rdr r 0 2 When you have to integrate over dA Pick a friendly coordinate system.. Use the appropriate dA Don’t forget the function s or may be functions of position. The coordinate system that you choose should match the symmetry of these two functions. An Example finite line of charge x dx r 1 dx dV 40 (d 2 x 2 )1/ 2 d 1 dx V 40 0 (d 2 x 2 )1/ 2 L P and 1 L ( L2 x 2 )1/ 2 V ln 40 d At P Using table of integrals Example (from text) z R disk s=charge per unit area s V 2 0 z 2 d z dz R z 2 V s Ez z 2 0 s z 1 Ez 2 2 2 0 z R 2 R2 z Which was the result we obtained earlier The Potential From a Dipole P r(+) + V ( P) Vi V () V () r r(-) d q - 1 q q V 40 r () r () q r ( ) r ( ) V 40 r ()r () Dipole - 2 P r(+) r + r(-) d q - Geometry r () r ( ) d cos(q ) r ( ) r ( ) r 2 q d cos(q ) 1 p cos(q ) V 2 40 r 40 r2 Where is this going? Charges and Forces Electric Fields Concept of Potential Batteries and Circuit Elements (R,C,L) Electric Circuits A Few Problems A particular 12 V car battery can send a total charge of 81 A · h (ampere-hours) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent? Sometimes you need to look things up … 1 ampere is 1 coulomb per second. 81 (coulombs/sec) hour = 81 x (C/s) x 3600 sec = 2.9 e +5 (b) If this entire charge undergoes a potential difference of 12 V, how much energy is involved? qV=2.9 e 05 x 12=3.5 e+6 An infinite nonconducting sheet has a surface charge density = 0.10 µC/m2 on one side. How far apart are equipotential surfaces whose potentials differ by 54 V? d 54V qEd 54volts s 0.1e 6 E 5.65e3 2 0 2 x8.85e 12 Ed V 54 3 d 9 x10 meters 5.65e3 In a given lightning flash, the potential difference between a cloud and the ground is 2.3x 109 V and the quantity of charge transferred is 43 C. (a) What is the change in energy of that transferred charge? (GJ) Energy = qV= 2.3 e+09 x 43C=98.9 GJ (b) If all the energy released by the transfer could be used to accelerate a 1000 kg automobile from rest, what would be the automobile's final speed? m/s E=(1/2)Mv2 v= sqr(2E/M)= 14,100 m/s Electrical Circuits Do Something! Include a power source Includes electrical components resistors capacitors inductors transistors diodes tubes switches wires Battery Maintains a potential difference between its two terminals. The potential Difference – the “voltage” is maintained by an electrochemical reaction which takes place inside of the battery. Has two terminals + is held at the higher potential - is held at the lower potential Symbol DC Power Source Low Potential Potential High SWITCH schematic symbol