Chapter 25 25 1 25 2 25 3

Chapter 25 25 -1 25 -2 25 -3 charges 11/29/2020 electric potential Potential difference and electric Potential Difference and electric field Electric Potential and Potential energy due to point Norah Ali Al-moneef king saud university 1

25 -1 Potential difference and electric Potential 11/29/2020 Norah Ali Al-moneef king saud university 2

Work and Potential Energy Electric Field Definition: Work Energy Theorem a b 11/29/2020 Norah Ali Al-moneef king saud university 3

Electric Potential Difference a b Definition: 11/29/2020 Norah Ali Al-moneef king saud university 4

Conventions for the potential “zero point” “Potential” 0 0 Choice 1: Va=0 0 0 Choice 2: 11/29/2020 Norah Ali Al-moneef king saud university 5

25 -2 Potential Difference and electric field When a force is “conservative” ie gravitational and the electrostatic force a potential energy can be defined Change in electric potential energy is negative of work done by electric force: ∆ V = -∫ E ds = -Ed 11/29/2020 Norah Ali Al-moneef king saud university 6

• The change in potential energy is directly related to the change in voltage. DU = q. DV DV = DU/q • DU: change in electrical potential energy (J) • q: charge moved (C) • DV: potential difference (V) • All charges will spontaneously go to lower potential energies if they are allowed to move. Units of Potential Difference Because of this, potential difference is often referred to as “voltage” In addition, 1 N/C = 1 V/m - we can interpret the electric field as a measure of the rate of change with position of the electric potential. So what is an electron Volt (e. V)? 11/29/2020 Norah Ali Al-moneef king saud university 7

Electron-Volts • Another unit of energy that is commonly used in atomic and nuclear physics is the electron-volt • One electron-volt is defined as the energy a chargefield system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt 1 e. V = 1. 60 x 10 -19 J • Since all charges try to decrease UE, and DUE = q. DV, this means that spontaneous movement of charges result in negative DU. • DV = DU / q • Positive charges like to DECREASE their potential (DV < 0) • Negative charges like to INCREASE their potential. (DV > 0) 11/29/2020 Norah Ali Al-moneef king saud university 8

VB – V A = V C - V A VB = V C A uniform electric field directed along the positive x axis. Point B is at a lower electric potential than point A. Points B and C are at the same electric potential. 11/29/2020 Norah Ali Al-moneef king saud university 9

Example If a 9 V battery has a charge of 46 C how much chemical energy does the battery have? E = V x Q = 9 V x 46 C = 414 Joules 11/29/2020 Norah Ali Al-moneef king saud university 10

Example A pair of oppositely charged, parallel plates are separated by 5. 33 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field strength between the plates? (b) What is the magnitude of the force on an electron between the plates? 11/29/2020 Norah Ali Al-moneef king saud university 11

Example Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V 11/29/2020 Norah Ali Al-moneef king saud university 12

25 -3 Electric Potential and Potential energy due to point charges +Q ds for a point charge 11/29/2020 Norah Ali Al-moneef king saud university 13

Recall the convention for the potential “zero point” Equipotential surfaces are concentric spheres 11/29/2020 Norah Ali Al-moneef king saud university 14

Superposition of potentials +Q 1 +Q 2 0 +Q 3 11/29/2020 Norah Ali Al-moneef king saud university 15

E and V for a Point Charge • The equipotential lines are the dashed blue lines • The electric field lines are the brown lines • The equipotential lines are everywhere perpendicular to the field lines An equipotential surface is a surface on which the electric potential is the same everywhere. 11/29/2020 Norah Ali Al-moneef king saud university 16

Figure 25. 4 (Quick Quiz 25. 3) Four equipotential surfaces Equipotential surfaces (the dashed blue lines are intersections of these surfaces with the page) and electric field lines (red- rown lines) for (a) a uniform electric field produced by an infinite sheet of charge, (b) a point charge, In all cases, the equipotential surfaces are perpendicular to the electric field lines at Norah Ali Al-moneef 11/29/2020 king saud university every point 17

Example (25. 1) A 12 -V battery connected to two parallel plates. The electric field between the plates has a magnitude given by the potential difference V divided by the plate separation d =0. 3 cm Example (25. 2) 11/29/2020 Norah Ali Al-moneef king saud university 18

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Example: (a) In figure a, 12 electrons are equally spaced and fixed around a circle of radius R. Relative to V=0 at infinity, what are the electric potential and electric field at the center C of the circle due to these electrons? (b) If the electrons are moved along the circle until they are nonuniformly spaced over a 120 are (figure b), what then is the potential at C? Solution: 11/29/2020 Norah Ali Al-moneef king saud university 20

Potential due to a group of point charges Example (25. 3) (a) The electric potential at P due to the two charges q 1 and q 2 is the algebraic sum of the potentials due to the individual charges. (b) A third charge q 3 = 3. 00 C is brought from infinity to a position near the other charges. 11/29/2020 Norah Ali Al-moneef king saud university 21

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Example An electric dipole consists of two charges q 1 = +12 n. C and q 2 = -12 n. C, placed 10 cm apart as shown in the figure. Compute the potential at points a, b, and c. 11/29/2020 Norah Ali Al-moneef king saud university 23

Example The Total Electric Potential At locations A and B, find the total electric potential.

(a) If two point charges are separated by a distance r 12, the potential energy of the pair of charges is given by keq 1 q 2/r 12. (b) If charge q 1 is removed, a potential keq 2/r 12 exists at point P due to charge q 2. 11/29/2020 Norah Ali Al-moneef king saud university 25

Potential energy due to multiple point charges +Q 2 +Q 1 +Q 3 11/29/2020 Norah Ali Al-moneef king saud university 26

Example 1. What is the potential energy if a +2 n. C charge moves from ¥ to point A, 8 cm away from a +6 C charge? The P. E. will be positive at point A, because the field can do + work if q is released. Potential Energy: A · 8 cm +2 n. C +Q +6 C U = 1. 35 m. J 11/29/2020 Positive potential Norah Ali Al-moneef energy king saud university 27

Signs for Potential Energy Consider Points A, B, and C. A · For +2 n. C at A: U = +1. 35 m. J · 12 cm 8 cm · C +Q Questions: If +2 n. C moves from A to B, does field E do + or – work? Does P. E. increase or decrease? B 4 cm +6 C Moving positive q +2 n. C The field E does positive work, the P. E. decreases. If +2 n. C moves from A to C (closer to +Q), the field E does negative work and P. E. increases. 11/29/2020 Norah Ali Al-moneef king saud university 28

Example. What is the change in potential energy if a +2 n. C charge moves from A to B? A · Potential Energy: 8 cm · B 12 cm +Q From Ex-1: UA = + 1. 35 m. J +6 C DU = UB – UA = 0. 9 m. J – 1. 35 m. J DU = -0. 450 m. J Note that P. E. has decreased as work is done by E. 11/29/2020 Norah Ali Al-moneef king saud university 29

Example What is the change in potential energy if a -2 n. C charge moves from A to B? A · Potential Energy: From Ex-1: UA = -1. 35 m. J 8 cm (Negative due to – charge) · B 12 cm +Q +6 C DU = +0. 450 m. J UB – UA = -0. 9 m. J – (-1. 35 m. J) A – charge moved away from a + charge gains P. E. 11/29/2020 Norah Ali Al-moneef king saud university 30

Example : Find the potential at a distance of 6 cm from a – 5 n. C charge. P. q = – 4 C r 6 cm - -Q - - Q = -5 n. C Negative V at Point P: VP = -750 V What would be the P. E. of a – 4 C charge placed at this point P? U = q. V = (-4 x 10 -6 C)(-750 V); U = 3. 00 m. J Since P. E. is positive, E will do + work if q is released. 11/29/2020 Norah Ali Al-moneef king saud university 31

Example : Two charges Q 1= +3 n. C and Q 2 = -5 n. C are separated by 8 cm. Calculate the electric potential at point A. B · 2 cm Q 1 + +3 n. C 6 cm A · 2 cm VA = 450 V – 2250 V; 11/29/2020 VA = -1800 V Norah Ali Al-moneef king saud university Q 2 = -5 n. C 32

Example Calculate the electric potential at point B for same charges. B · 2 cm Q 1 + +3 n. C 6 cm A · 2 cm VB = 1350 V – 450 V; 11/29/2020 VB = +900 V Norah Ali Al-moneef king saud university Q 2 = -5 n. C 33

Example : What is the potential difference between points A and B. What work is done by the E-field if a +2 C charge is moved from A to B? VA = -1800 V VB = +900 V VBA= VB– VA = 900 V – (-1800) V VAB = +2700 V Note point B is at higher potential. d. WAB = -d. U= q(VA – VB) = -(2 x 10 -6 C )(+2700 V) Work = -5. 40 m. J B · Q 1 + 2 cm +3 n. C 6 cm A Q 2 · - 2 cm -5 n. C E-field does negative work. du increased Thus, an external force was required to move the charge. 11/29/2020 Norah Ali Al-moneef king saud university 34

Example 6 (Cont. ): Now suppose the +2 C charge is moved from back from B to A? VA = -1800 V VB = +900 V VAB= VA– VB = -1800 V – 900 V B · Q 1 + 2 cm +3 n. C 6 cm VBA = -2700 V This path is from high to low potential. d. WAB =-d. U= -q(VA – VB) = -(2 x Work = +5. 40 m. J 10 -6 C )(-2700 V) A Q 2 · - 2 cm -5 n. C E-field does positive work. d. U decreased so The work is done BY the E-field this time ! 11/29/2020 Norah Ali Al-moneef king saud university 35

Example An electron is accelerated in a TV tube through a potential difference of 5000 V. a) What is the change in PE of the electron? V = DPE/q DPE = q. V = (-1. 60 X 10 -19 C)(+5000 V)= -8. 0 X 10 -16 J What is the final speed of the electron (m = 9. 1 X 10 -31 kg) DPE + DKE = 0 (Law of conservation of energy) DPE = -DKE DPE = - ½ mv 2 = (-2)(DPE) = (-2)(-8. 0 X 10 -16 J) m 9. 1 X 10 -31 kg v = 4. 2 X 107 m/s 11/29/2020 Norah Ali Al-moneef king saud university 36

Summary • Electric potential energy: • Electric potential difference: work done to move charge from one point to another • Relationship between potential difference and field: • Equipotential: line or surface along which potential is the same • Electric potential of a point charge: 11/29/2020 Norah Ali Al-moneef king saud university 37

1: 2: 11/29/2020 Norah Ali Al-moneef king saud university 38

3: 4: 11/29/2020 Norah Ali Al-moneef king saud university 39

5: The electrons in a particle beam each have a kinetic energy of 1. 60 x 10 -17 J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10. 0 cm? 6: An electron and a proton are each placed at rest in an electric field of 520 N/C. Calculate the speed of each particle 48. 0 ns after being released. 11/29/2020 Norah Ali Al-moneef king saud university 40

q 1 d 7: What is the potential at point P, located at the center of the square of point charges. Assume that d = 1. 3 m and the charges are q 1 = +12 n C, q 3 = +31 n C, 11/29/2020 q 2= -24 n C q 4= +17 n C Norah Ali Al-moneef king saud university d P d d q 3 q 4 41

1 - The electric field has a magnitude of 3. 0 N/m at a distance of 60 cm from a point charge. What is the charge? (a) 1. 4 n. C (b) 120 p. C (c) 36 m. C (d) 12 C (e) 3. 0 n. C 11/29/2020 Norah Ali Al-moneef king saud university 42

1 - A conducting sphere has a net charge of − 4. 8 × 10− 17 C. What is the approximate number of excess electrons on the sphere? (a) 100 (b) 200 (c) 300 (d) 400 (e) 500 Electric charge always occurs in multiples of e Q = Ne e = 1. 60 × 10 19 (N =1、2、3…) C N= (-4. 8 x 10 -17 C/-1. 6 x 10 -19 C=300 electrons) 2 - Two point charges, 8 x 10 -9 C and -2 x 10 -9 C are separated by 4 m. The electric field magnitude (in units of V/m) midway between them is: A) 9 x 109 B) 13, 500 C) 135, 000 D) 36 x 10 -9 E) 22. 5 11/29/2020 Norah Ali Al-moneef king saud university 43

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3 - If 10000 electrons are removed from a neutral ball, its charge is; (a) +1. 6× 10 -15 C (b) +1. 6× 10 -23 C (c) -1. 6× 10 -15 C (d) -1. 6× 10 -23 C Q = Ne =10000 x -1. 6× 10 -19 Q = -1. 6× 10 -15 C 4 - A charge of 10 -6 C is in a field of 9000 N/C, directed upwards. The magnitude and direction of the force it experiences are; (a) 9× 10 -3 N, downwards (b) 3× 10 -3 N, downwards (c) 9× 10 -3 N, upwards (d) 3× 10 -3 N, upwards F= q E = 9000 x 10 -6 F = 9 x 10 -3 N 11/29/2020 Norah Ali Al-moneef king saud university 45