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696_PartUniversity Physics Solution

# 696_PartUniversity Physics Solution - Electric Potential...

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Electric Potential 23-19 EVALUATE: Notice that an increasing potential does not necessarily mean that the electric field is increasing. Figure 23.52 23.53. (a) IDENTIFY: Apply the work-energy theorem, Eq.(6.6). SET UP: Points a and b are shown in Figure 23.53a. Figure 23.53a EXECUTE: 5 tot 4.35 10 J bab WK K K K =Δ = = = × The electric force E F and the additional force F both do work, so that tot . E F F WWW =+ 55 5 tot 4.35 10 J 6.50 10 J 2.15 10 J E FF −− =− = × × = −× EVALUATE: The forces on the charged particle are shown in Figure 23.53b. Figure 23.53b The electric force is to the left (in the direction of the electric field since the particle has positive charge). The displacement is to the right, so the electric force does negative work. The additional force F is in the direction of the displacement, so it does positive work. (b) IDENTIFY and SET UP: For the work done by the electric force, ( ) ab a b Wq V V EXECUTE: 5 3 9 2.15 10 J 2.83 10 V. 7.60 10 C W VV q −= = = − × × EVALUATE: The starting point (point a ) is at 3 2.83 10 V × lower potential than the ending point (point b ). We know that ba > because the electric field always points from high potential toward low potential. (c) IDENTIFY: Calculate E from and the separation d between the two points. SET UP: Since the electric field is uniform and directed opposite to the displacement , E WF d q E d where 8.00 cm d = is the displacement of the particle. EXECUTE: 3 4 3.54 10 V/m. 0.0800 m a b WV V E qd d × = = × EVALUATE: In part (a), tot W is the total work done by both forces. In parts (b) and (c) W is the work done just by the electric force. 23.54. IDENTIFY: The electric force between the electron and proton is attractive and has magnitude 2 2 ke F r = . For circular motion the acceleration is 2 rad / av r = . 2 e Uk r . SET UP: 19 1.60 10 C e . 19 1 eV 1.60 10 J . EXECUTE: (a) 22 2 mv ke rr = and 2 ke v mr = . (b) 2 2 11 1 2 ke K mv U r ===

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23-20 Chapter 23 (c) 21 9 2 18 11 1 1 1 (1.60 10 C) 2.17 10 J 13.6 eV 22 2 5 . 2 9 1 0 m ke k EKU U r × =+= = = = − × = × . EVALUATE: The total energy is negative, so the electron is bound to the proton. Work must be done on the electron to take it far from the proton. 23.55. IDENTIFY and SET UP: Calculate the components of E G from Eq.(23.19). Eq.(21.3) gives F G from E. G EXECUTE: (a) 4/3 VC x = 3 4 / 240 V/(13.0 10 m) 7.85 10 V/m CVx == × = × (b) 1/3 5 4 (1.05 10 V/m ) 3 x V E Cx x x =− × The minus sign means that x E is in the x -direction, which says that E G points from the positive anode toward the negative cathode. (c) q F=E GG so 1/3 4 3 xx Fe Ee C x = Halfway between the electrodes means 3 6.50 10 m. x 19 4 3 15 4 3 (1.602 10 C)(7.85 10 V/m )(6.50 10 m) 3.13 10 N x F −− × × = × x F is positive, so the force is directed toward the positive anode. EVALUATE: V depends only on x, so 0. yz EE = = E G is directed from high potential (anode) to low potential (cathode). The electron has negative charge, so the force on it is directed opposite to the electric field.
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696_PartUniversity Physics Solution - Electric Potential...

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