Chapter 22 : Electric potential

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1 Chapter 22 : Electric potential What is electric potential? How does it relate to potential energy? How does it relate to electric field? Some simple applications

2 What does it mean when it says 1.5 Volts on the battery? The electric potential difference between the ends is 1.5 Volts

3 230 V 1.5 V 100,000 V So what is a volt?

4 The electric potential difference VV in volts between two points is the work in Joules needed to move 1 C of charge between those points WW = qq VV W = work done [in J] q = charge [in C] V = potential difference [in V] VV is measured in volts [V] : 1 V = 1 J/C

5 The electric potential difference VV in volts between two points is the work in Joules needed to move 1 C of charge between those points WW = qq VV The 1.5 V battery does 1.5 J of work for every 1 C of charge flowing round the circuit

6 Potential energy What is this thing called potential? Potential energy crops up everywhere in physics

7 Potential energy Potential energy U is the energy stored in a system (when work is done against a force) e.g. force of gravity FF = mmmm Work = Force x Distance FF h WW = FF h = mmmmm UU = mmmmm

8 Potential energy Potential energy may be released and converted into other forms (such as kinetic energy) Work is done, increasing the potential energy

9 Potential energy Potential energy difference is the only thing that matters not the reference (or zero) level For example, applying conservation of energy to a mechanics problem: Final energy = Initial energy KKKK ffffffffff + PPPP ffffffffff = KKKK iiiiiiiiiiiiii + PPPP iiiiiiiiiiiiii KKKK ffffffffff = KKKK iiiiiiiiiiiiii + (PPPP iiiiiiiiiiiiii PPPP ffffffffff ) Difference in potential energy

10 Potential energy Potential energy difference doesn t depend on the path only on the two points A and B

11 Potential energy Potential energy U is the energy stored in a system second example e.g. stretching a spring FF FF = kkkk Work = Force x Distance Force is varying with distance! WW = kkkk dddd = 1 2 kkxx2 xx UU = 1 2 kkxx2

12 e.g. moving a charge through an electric field EE qq xx FF FF = qqqq (minus sign because the force is opposite to E) Work = Force x Distance WW = FF xx = qqqq xx Potential difference VV is work needed to move 1C of charge: WW = qq VV Equate: qq VV = qqqq xx EE = VV xx

13 Electric field is the gradient of potential EE = VV High V EE Low V VV xx Positive charges feel a force from high to low potential xx xx Negative charges feel a force from low to high potential

14 Two parallel plates have equal and opposite charge. Rank the indicated positions from highest to lowest electric potential A B D C A=C, B=D 2. A, B, C, D 3. C, D=B, A 4. A, B=D, C 0% 0% 0% 0%

15 Analogy with gravitational potential Gravitational potential difference exerts force on mass Electric potential difference exerts force on charge VV qq xx

16 Electric field is the gradient of potential EE = VV High V EE Low V xx The dashed lines are called equipotentials (lines of constant V) Electric field lines are perpendicular to equipotentials It takes no work to move a charge along an equipotential (work done = dddd = FF. dddd = qqee. dddd = 0)

17 Summary for two plates at potential difference V EE Electric field is the potential gradient EE = VV dd Work W to move charge q from ve to +ve plate dd WW = qq VV

18 Link to potential energy The electric potential difference VV between two points is the work needed to move 1 C of charge between those points WW = qq VV This work is also equal to the potential energy difference UU between those points UU = qq VV Potential V = potential energy per unit charge U/q

19 An electron is placed at X on the negative plate of a pair of charged parallel plates. For the maximum work to be done on it, which point should it be moved to? 1. A 2. B 3. C 4. D 5. A or C 6. C or D X B A D C % 0% 0% 0% 0% 0%

20 What is the electric potential near a charge +Q? Work = Force x Distance FF = kk QQ qq +q xx 2 Force is varying with distance, need integral! rr rr kk QQ qq WW = FF dddd = xx 2 dddd = kk QQ qq rr xx +Q Potential energy UU = kk QQ qq rr Electric potential VV = UU qq = kk QQ rr

21 What is the electric potential near a charge +Q? +q rr +Q Electric potential VV = kk QQ rr

22 Exercise: a potential difference of 200 V is applied across a pair of parallel plates m apart. (a) calculate E and draw its direction between the plates. The electric field is the gradient in potential EE = VV xx = = VV mm 1 [oooo NN CC 1 ] VV = 200 VV = 0 EE +ve plate -ve plate

23 Exercise: a potential difference of 200 V is applied across a pair of parallel plates m apart. (b) an electron is placed between the plates, next to the negative plate. Calculate the force on the electron, the acceleration of the electron, and the time it takes to reach the other plate. Force FF = qqqq = ( ) ( ) = NN FF = mmmm Acceleration aa = FF mm EE ee +ve plate -ve plate = = mm ss 2 dd = 1 2 aatt2 Time tt = 2dd aa tt = = ss e = 1.6 x C; m e = 9.1 x kg

24 Exercise: a potential difference of 200 V is applied across a pair of parallel plates m apart. (c) calculate the work done on the electron as it travels between the plates. The potential difference is the work done on 1C charge Work WW = qqqq = = JJ EE ee +ve plate -ve plate e = 1.6 x C; m e = 9.1 x kg

25 Chapter 22 summary Electric potential difference V is the work done when moving unit charge: WW = qqqq The electric potential energy is therefore also given by: UU = qqqq The electric field is the gradient of the potential: EE = VV/ xx Charges feel a force from high electric potential to low potential

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EE 471: Transport Phenomena in Solid State Devices Spring 2018 Lecture 2 Electrons and Holes in Semiconductors Bryan Ackland Department of Electrical and Computer Engineering Stevens Institute of Technology

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Interaction with matter accelerated motion: ss = bb 2 tt2 tt = 2 ss bb vv = vv 0 bb tt = vv 0 2 ss bb EE = 1 2 mmvv2 dddd dddd = mm vv 0 2 ss bb 1 bb eeeeeeeeeeee llllllll bbbbbbbbbbbbbb dddddddddddddddd

(b) The type of matter is irrelevant since the energy is directly proportional to mass only

Exercise J.3.1. Answers 1. m = 800kg v = 70kmh -1 = 70 103 = 60 60 19.4ms-1 KK. EE = 1 2 mmvv2 = 800 19.42 2 = 150544JJ Using E= mc 2 and the kinetic energy of the car we obtain mm = EE cc 2 = 151235 (3