Chapter_20_Worked Out Practice Problems 3. Picture the Problem: A uniform electric field E = 6.8 × 105 N/C( )ˆx creates a change in the electric potential. Strateg y: The change in electric potential is given by equation 20-4, ΔV = − EΔs , where Δs is equal to Δx because the field points in the ˆxdirection. There is no change in the electric potential along the ˆy and ˆzdirections. The change in electric potential from point A to point B is independent of the path taken between those two points. Solution: 1. (a) Here Δx = 0because the path is in the ˆydirection: ΔV = 0 2. (b) Solve equation 20-4 for ΔV : ΔV = − E Δx = − 6.8 × 105N/C( )6.0 m( )= − 4.1 × 106 V 3. (c) The potential changes by an amount equal to the sum of paths 1 and 2 above: ΔV = 0 − 4.1 × 106 V = − 4.1 × 106 V = − 4.1 MV Insight: Note that the potential changes only along the direction parallel to the field. This is consistent with statement, “The electric field points downhill on the potential surface,” as described by figure 20-3 and the accompanying text. 4. Picture the Problem: The cell membrane acts like a parallel plate capacitor, which is modeled in the diagram shown at right. Strateg y: The field across the cell membrane is uniform, like that between the plates of a parallel plate capacitor. Use equation 20-4 to find the magnitude of the field that corresponds to the potential difference between the plates. Solution: 1. Apply equation 20-4 directly: E =ΔVΔs=ΔVd=0.070 V0.10 × 10− 6 m= 7.0 × 105 V m 2. The electric field points from higher to lower potential, so the electric field is directed inward across the cell membrane. Insight: The cell membrane was also modeled as a parallel plate capacitor in problems 52, 68, and 74 of Chapter 19. 8. Picture the Problem: An ion loses electric potential energy as it is accelerated through a potential difference. Strategy: The electric potential is the electric potential energy per charge (equation 20-2). To find the charge we need only divide the electric potential energy by the electric potential. Solution: Solve equation 20-2 for q0: q0=ΔUΔV=−1.37 × 10−15 J2140 V= − 6.40 × 10−19C = −4.00 e Insight: If the charge were positive, its electric potential energy would increase as it moves from a lower to a higher potential. In this problem the negative charge loses electric potential energy and gains kinetic energy as it accelerates. 19. Picture the Problem: An electron accelerates through a large potential difference in a TV picture tube. Strategy: The electron will accelerate toward a higher electric potential due to its negative charge. The change in potential energy is the charge times the potential difference (equation 20-2). The change in potential energy equals the gain in kinetic energy, which can then be used to find the speed. Solution: 1. Set the kinetic energy equal to the change in potential energy and solve for v: 12mv2= ΔU = eΔV ⇒ v =2eΔVm2. Find the speed of the electron: v =2eΔVm=2 1.60 × 10−19C( )25,000 V( )9.11× 10−31kg= 9.4 × 107 m/s Insight: This speed is about 30% of the speed of light, so that relativistic effects are becoming important (see Chapter 29). When relativistic effects are taken into consideration, we find the speed is actually 9.0×107 m/s. 24. Picture the Problem: For the figure shown at right, it is given that q1 = +Q. Strategy: Use the expression for the potential due to a point charge (equation 20-7), together with the knowledge that the potentials due to different point charges simply add, to determine the value of q2 and the potential at point A. Solution: 1. (a) The potential at point B is the sum of the potentials due to q1 and q2. Each potential is given by V = k q r(equation 20-7). The distance r from q1 to the point B is 2 times greater than the distance from q2 to the point B. Therefore, for the electric potential to be zero at point B, it is necessary that q2 have the charge − Q 2 . 2. (b) The potential at point A is the sum of the potentials due to q1 and q2. The charge q1 is both larger in magnitude and closer to point A, and we conclude that the electric potential at point A will be positive. Insight: If the square has side length a we can calculate: VA=k Qa+k − Q 2( )2 a=k Qa1 −12"#$%&'=k Q2a. 29. Picture the Problem: Two charges are located in space a short distance from each other. Strategy: The electric potential energy for a pair of charges is the charge on one of them multiplied by the electric potential created by the other, or U = q1V2= k q1q2r. Solve this expression for r to find the required distance. Solution: Solve U = k q1q2r for r: r =kq1q2U=8.99 × 109N ⋅ m2/ C2( )7.22 × 10− 6C( )−26.1 × 10− 6C( )−126 J= 0.0134 m = 1.34 cm Insight: The electric potential energy associated with a system of three charges is discussed in active example 20-3. 38. Picture the Problem: Three charges are arranged at the corners of a rectangle as indicated in the diagram at right. Strategy: The work required to move all three charges to infinity is equal to the change in the total electric potential energy of all three charges. This energy is the sum of terms like U = kq1q2r for each pair of charges in the system, as illustrated in active example 20-3. Sum the potential energies for each of the three pairs of charges in this system to find the work required to move all of them infinitely far from one another. Let r23= 0.25m( )2+ 0.16 m( )2= 0.30 m. Solution: 1. Find ΔUby summing the terms from each pair of charges: W = ΔU = 0 −kq1q2r12+kq1q3r13+kq2q3r23#$%&'(= −kq1q2r12+q1q3r13+q2q3r23#$%&'( 2. Factor out 10−12C2 from each product of charges and calculate W: W = − 8990 N ⋅m2( )− 6.1( )2.7( )0.25 m+− 6.1( )−3.3( )0.16 m#$%%+2.7( )−3.3( )0.30 m&'((= − 0.27 J Insight: Note that the work done by the electric field is W = −ΔU(equation 20-1), but the work you must do to move the charges against the field is W = ΔU . The negative result means the charges would fly off on their own. 41. Picture the Problem: Four charges of the same magnitude but different signs are arranged at the corners of a square of side length a as shown at right.Strategy: The total electric potential energy is the sum of terms like U = kq1q2r for each pair of charges in the system, as illustrated by active example 20-3.
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