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Transcript
Electric Potential, Energy, and Capacitance Objective: TSW understand, transfer and apply energy concepts to electric fields and charges by solving problems involving electric fields and forces. Let’s revisit energy concepts for a gravitational field. A 250 gram baseball is thrown upward with an initial velocity of 25m/s. What is the maximum height reached by the ball? v = 0m/s E0 W E f K Ug 1 2 mv mgh 2 2 v h 2g v = 25m/s EARTH (25) 2 h 2(10) h 31.3m Let’s make a parallel comparison of Gravitational Potential Energy (Ug) to Electric Potential Energy (UE). U g mgh Units: U E qEd Units: N kg m Nm J kg N C m Nm J C UE = Energy that a charge has due to its position in an electric field. Now we will do an example: An electron is released from rest in an electric field of 2000N/C. How fast will the electron be moving after traveling 30cm? E0 W E f v=? _ v = 0m/s _ 30cm UE K 1 2 qEd mv 2 2qEd v m 2(1.6 1019 )( 2000)(.3) v 9.111031 m v 1.45 10 s 7 A new quantity is defined called the Electric Potential Difference. It is the electric potential energy difference per unit charge between two points in an electric field. Electric Potential Energy Electric Potential Difference Charge UE V q qEd V q V Ed The Electric Potential Difference can also called the Electric Potential, the Potential, or the voltage. Units: J volt V C Remember Energy and voltage are scalars, so you don’t have to deal with vectors (direction) The electric potential energy can now be written in terms of the electric potential. U E qEd V Ed U E qV Check the units qV = (C)(J/C) = J Let’s do an example using this new concept: The potential difference between two charge plates is 500V. Find the velocity of a proton if it is accelerated from rest from one plate to the other. E W E 0 High Low Potential + Potential - + - + + - + - + - 500V Positive charges move from high to low potential Negative charges move from low to high potential f UE K 1 2 qV mv 2 2qV v m 2(1.6 10 19 )(500) v 1.67 10 27 5 m v 3.110 s Let’s revisit a gravitational field again: The gravitational potential energy if the field is constant is given by: U g mgh But, what if the field is not constant. In other words the gravitational potential energy of a mass located far away from the earth. m r The gravitational potential energy of mass m is given by: M Em U g G r Notice r is not squared The gravitational potential energy at infinity is zero Let’s use this new equation to do an example: With what velocity must a rocket be fired in order to escape the earth’s gravitational field? (Neglect air resistance) E0 W E f K Ug 0 1 2 M Em mv G 0 2 r 1 2 M Em 11 24 mv G 2 ( 6 . 67 10 )( 5 . 98 10 ) 2 r v 6.38 106 2GM E v m v 11000 r s Let’s make a parallel comparison of Gravitational Potential Energy (Ug) to Electric Potential Energy (UE) when dealing with individual masses or charges. The gravitational potential energy between two masses is given by: M Em U g G r The electric potential energy between two masses is given by: q1q2 UE k r Example: E0 W E f U E K1 K 2 q1q2 1 2 1 2 k mv mv r 2 2 q2 k mv 2 r 2 kq v2 rm Two 40 gram masses each with a charge of -6µC are 20cm apart. If the two charges are released, how fast will they be moving when they are a very, very long way apart. (infinity) kq2 (9 109 )(6 10 6 ) 2 m v 6.36 rm (.2)(. 04) s Let’s use the electric potential energy between two charges to derive an equation for the electric potential (voltage) due to a single point charge. The electric potential energy is given by: q2P r q1 q1q2 UE k r Let’s remove charge q2 and consider the electric potential (voltage) at point P. kq1q2 UE q1 r V k q q2 r To find the potential due to more than one point charge simply add up all the individual potentials: i q V k r i Example 1: The electron in the Bohr model of the atom can exist at only certain orbits. The smallest has a radius of .0529nm, and the next level has a radius of .212nm. a) What is the potential difference between the two levels? b) Which level has a higher potential? q V k r e V1 k r1 r1 r2 19 1.6 10 V1 (9 10 ) 27.2V 9 .0529 10 19 1 . 6 10 V2 (9 109 ) 6.79V 9 .212 10 +e 9 r1 is at a higher potential. potential diff V 27.2 6.79 20.4V Example 2 What is the electric potential at the center of the square? 45º 45º r r r 2k V V qq2 kr rr 0.10 6 6 10 5 2 r 9 10 10 9.01 10 V V 9 910 071 . . 071 m r .071 65 J 10 3410 V V 16..27 C 2 r r V k i 2 qi r Vtotal 1.27 106 (1.27 106 ) 6.34 105 6.34 105 J Vtotal 1.27 10 C 6 Example 3: A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C. a) How much work would be required to move a proton from the negative to the positive plate? b) What is the potential difference between the plates? c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate? W U E qEd N W (1.6 10 C )(1500 )(.015m) C W 3.6 10 18 J 19 Example 3: A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C. b) What is the potential difference between the plates? V Ed N V (1500 )(.015m) C J V 22.5 C Example 3: A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C. c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate? Use conservation of energy E0 W E f UE K 1 2 qV mv 2 2qV v m 2(1.6 10 19 )( 22.5) v 1.67 10 27 m v 6.57 10 4 s Example 4: Compute the energy necessary to bring together the charges in the configuration shown below: Calculate the electric potential energy between each pair of charges and add them together. qq12q1qq332 UU13 2312kk rr 66 6 666 ( 4 10 )( 4 10 ( 4 10 )( 4 10 )) 99 9 UU13 (910 10 )) 2312(9 .2 UU13 72JJ 231200..72 U total .72 J (.72 J ) (.72 J ) U total .72 J Capacitors Two parallel conducting plates used to store charge, which in turn stores energy Charge flows through the circuit until the capacitor is charged at which point no more charge flows. Capacitance The amount of charge stored per volt of potential difference. Q C V C = Capacitance (C/V = Farad (F)). Q = The stored charge (C). V = The potential difference (voltage) between the plates (V). Capacitance based on the physical characteristics of the capacitor C 0 A d Є0 = Permittivity of free space = 8.85 x 10-12 A = Surface area of one plate. d = distance between plates. Example 1: Find the Charge stored in the capacitor. Q C V Q CV 12V 20μF C Q (20 10 )(12V ) V Q 2.4 10 4 C 6 Capacitors in Series 1 1 1 1 ........ Cs C1 C2 C3 1 1 Cs i Ci Example 2: Find the total charge stored in each the capacitor. FirstNow calculate thetotal totalcharge capacitance. find the 5μF 12V 10μF 30μF 1Q 1 C CsV i Ci Q 1CV 1 1 1 6 Q C(s 3.051010F )(30 12V ) 5 C 3 . 0 Q s3.6 10 FC In a series circuit the charge stored on each capacitor is the same and the voltage is split. Capacitors in Parallel C p C1 C2 C3 ...... C p Ci i Example 3: Find the total charge stored in all the capacitors. Now find calculate the First the total total charge. capacitance. 12V 5μF 10μF 30μF Q CC p V Q CV C i i C p 5 10 30 Q (45 10 6 F )(12V ) C p 454F Q 5.4 10 C In a parallel circuit the charge stored on each capacitor can be different and the voltage must be the same. Example 4: Find the total capacitance of all the capacitors. First find the total capacitance of the parallel capacitors 12μF C p 8 4 12F 12V 8μF 4μF Now, combine this value in series with 12μF capacitor. 1 Ctotal 1 1 12 12 Ctotal 6 F Energy Stored in a Capacitor 2 1 Q 1 2 U C QV CV 2 2C 2 Units: Joules (J)