Chapter 25 Electric potential 25 1 Potential difference
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Chapter 25 : Electric potential 25 -1 Potential difference and electric Potential 25 -2 Potential Difference and electric field 25 -3 charges Electric Potential and Potential energy due to point 10/30/2021 Norah Ali Al-moneef king saud university 1
25 -1 Potential difference and electric Potential 10/30/2021 Norah Ali Al-moneef king saud university 2
Work and Potential Energy Work Energy Theorem a b 10/30/2021 Norah Ali Al-moneef king saud university 3
Electric Potential Difference a b Definition: 10/30/2021 Norah Ali Al-moneef king saud university 4
Conventions for the potential “zero point” 0 0 “Potential” If Va=0 10/30/2021 Norah Ali Al-moneef king saud university 5
25 -2 Potential Difference and electric field Change in electric potential energy is negative of work done by electric force: 10/30/2021 Norah Ali Al-moneef king saud university 6
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Unit of Electric Potential 10/30/2021 Norah Ali Al-moneef king saud university 8
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 10/30/2021 Norah Ali Al-moneef king saud university 9
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? 10/30/2021 Norah Ali Al-moneef king saud university 10
Example Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V 10/30/2021 Norah Ali Al-moneef king saud university 11
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Superposition of potentials +Q 1 +Q 2 0 +Q 3 10/30/2021 Norah Ali Al-moneef king saud university 14
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Example (25. 2) 10/30/2021 Norah Ali Al-moneef king saud university 18
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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. 10/30/2021 Norah Ali Al-moneef king saud university 20
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. 10/30/2021 Norah Ali Al-moneef king saud university 22
Potential energy due to multiple point charges +Q 2 +Q 1 +Q 3 10/30/2021 Norah Ali Al-moneef king saud university 23
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 10/30/2021 Positive potential Norah Ali Al-moneef energy king saud university 24
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. 10/30/2021 Norah Ali Al-moneef king saud university 25
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. 10/30/2021 Norah Ali Al-moneef king saud university 26
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. 10/30/2021 Norah Ali Al-moneef king saud university 27
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. 10/30/2021 Norah Ali Al-moneef king saud university 28
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; 10/30/2021 VA = -1800 V Norah Ali Al-moneef king saud university Q 2 = -5 n. C 29
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; 10/30/2021 VB = +900 V Norah Ali Al-moneef king saud university Q 2 = -5 n. C 30
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 VAB= VA – VB = -1800 V – 900 V VAB = -2700 V Note point B is at higher potential. Work. AB = 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. Thus, an external force was required to move the charge. 10/30/2021 Norah Ali Al-moneef king saud university 31
Example 6 (Cont. ): Now suppose the +2 C charge is moved from back from B to A? VA = -1800 V VB = +900 V VBA= VB – VA = 900 V – (-1800 V) B · Q 1 + 2 cm +3 n. C 6 cm VBA = +2700 V This path is from high to low potential. Work. BA = q(VB – VA) = (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. The work is done BY the E-field this time ! 10/30/2021 Norah Ali Al-moneef king saud university 32
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 10/30/2021 Norah Ali Al-moneef king saud university 33
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: 10/30/2021 Norah Ali Al-moneef king saud university 34
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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. 10/30/2021 Norah Ali Al-moneef king saud university 37
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, 10/30/2021 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 38
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 10/30/2021 Norah Ali Al-moneef king saud university 39
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 10/30/2021 Norah Ali Al-moneef king saud university 40
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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 10/30/2021 Norah Ali Al-moneef king saud university 42
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