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Unformatted text preview: CHAPTER 25—ELECTRIC POTENTIAL ActivPhysics can help with these problems: Activities 11.9, 11.10 Section 25 2:—Potential Difference Problem 1. How much work does it take to move a 50 C μ charge against a 12V potential difference? Solution The potential difference and the work per unit charge, done by an external agent, are equal in magnitude, so W q V = = = C V J. ∆ ( )( ) 50 12 600 μ μ (Note: Since only magnitudes are needed in this problem, we omitted the subscripts A and B .) Problem 2. The potential difference between the two sides of an ordinary electrical outlet is 120 V. How much energy does an electron gain when it moves from one side to the other? Solution Moving from the negative to the positive side (i.e., opposite to the electric field), an electron gains e V eV ∆ = = 120 ( . )( ) . 16 10 120 192 10 19 17 × = × C V J of energy. Problem 3. It takes 45 J to move a 15mC charge from point A to point B . What is the potential difference ∆ V AB ? Solution The work done by an external agent equals the potential energy change, ∆ ∆ U q V AB AB = = 45 J , hence ∆ V AB = 45 15 3 J mC kV. = = (Since the work required to move the charge from A to B is positive, V V V B A AB and ∆ is positive.) Problem 4. Show that 1 V/m is the same as 1 N/C. Solution Since a volt joule/coulomb newtonmeter/coulomb, = = it follows that a V/m N/C = . Problem 5. Find the magnitude of the potential difference between two points located 1.4 m apart in a uniform 650 N/C electric field, if a line between the points is parallel to the field. Solution For in the direction of a uniform electric field, Equation 252b gives ∆ V E = = = ( )( .4 ) 650 1 910 N/C m V. (See note in solution to Problem 1. Since dV = ⋅ E d , the potential always decreases in the direction of the electric field.) Problem 6. A charge of 31 . C moves from the positive to the negative terminal of a 9.0V battery. How much energy does the battery impart to the charge? Solution ∆ ∆ U q V AB AB = = = C V J. ( . )( . ) . 31 9 0 27 9 2 CHAPTER 25 Problem 7. Two points A and B lie 15 cm apart in a uniform electric field, with the path AB parallel to the field. If the potential difference ∆ V AB is 840 V, what is the field strength? Solution Equation 252b for a uniform field gives E V = = = ∆ = = 840 5 60 V 0.15 m kV/m. . (See notes in solutions to Problems 1 and 5.) Problem 8. Figure 2537 shows a uniform electric field of magnitude E . Find expressions for (a) the potential difference ∆ V AB and (b) ∆ V BC . (c) Use your result to determine ∆ V AC . FIGURE 2537 Problem 8. Solution (a) On the line A to B , d is antiparallel to E , so Equation 252 gives V V E d Ed B A A B A B =  z ⋅ = z = E d . (b) The line B to C makes an angle of 45 ° with E , so V V E d Ed C B B C =  z ° =  cos . 45 2 = (c) Addition yields V V C A = V V V V Ed Ed C B B A + = = ( ) . . 1 1 2 0 293 = Problem 9. A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a...
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 Spring '09
 staff
 Charge, Electric Potential, Electrostatics, Work, Potential difference, Electric charge, NC, Vab

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