Physics Solution Manual for 1100 and 2101

110 a 0067 a 600 s 26 c b according to equation 194

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Unformatted text preview: aller = Vlarger dlarger Thus, we find that the voltage increases to a value of d larger Vlarger = Vsmaller d smaller 55. SSM REASONING 2dsmaller = ( 9.0 V ) d smaller = 18 V According to Equation 19.10, the capacitance of a parallel plate capacitor filled with a dielectric is C = κε 0 A / d , where κ is the dielectric constant, A is the area of one plate, and d is the distance between the plates. From the definition of capacitance (Equation 19.8), q = CV . Thus, using Equation 19.10, we see that the charge q on a parallel plate capacitor that contains a dielectric is given by q = (κε 0 A / d ) V . Since each dielectric occupies one-half of the volume between the plates, the area of each plate in contact with each material is A/2. Thus, q1 = κ1ε 0 ( A / 2) d V= κ1ε 0 A 2d q2 = and V κ 2ε 0 ( A / 2) d V= κ 2ε 0 A 2d V According to the problem statement, the total charge stored by the capacitor is q1 + q2 = CV (1) where q1 and q2 are the charges on the plates in contact with dielectrics 1 and 2, respectively. Using the expressions for q1 and q2 above, Equation (1) becomes CV = κ1ε 0 A 2d V+ κ 2ε 0 A 2d This expression can be solved for C. V= κ1ε 0 A + κ 2ε 0 A 2d V= (κ1 + κ 2 )ε 0 A 2d V Chapter 19 Problems SOLUTION Solving for C, we obtain C = 1045 ε 0 A(κ1 + κ 2 ) 2d 56. REASONING The electric potential energy stored in the capacitor is given by Energy = CV (Equation 19.11b), where C is the capacitance of the capacitor. We choose Equation 19.11b to express the energy because it contains the potential difference V across the plates of the capacitor. This quantity is related to the magnitude E of the electric field V between the capacitor’s plates via E = (Equation 19.7b), where d is the distance between d the plates. We will use Equations 19.11b and 19.7b to obtain an expression for the energy stored in the capacitor in terms of the magnitude E of the electric field between its plates κε A and its capacitance C. The capacitance itself is given by C = 0 (Equation 19.10), where d κ is the dielectric constant of the material between the plates, ε0 = 8.85×10−12 C2/(N·m2) is the permittivity of free space, and A is the surface area of one plate. The dielectric constant for air is, to two significant figures, κair = 1.0, and that for neoprene rubber is κneo = 6.7 (see Table 19.1). We note that the plate area A and the plate separation d are not given, but are identical for both capacitors. 1 2 2 V (Equation 19.7b) for V yields V = Ed. Substituting this into d (Equation 19.11b), we obtain SOLUTION Solving E = Energy = 1 CV 2 2 Energy = 1 CV 2 = 1 C ( Ed ) = 1 CE 2 d 2 2 2 2 2 Substituting C = κε 0 A d (1) (Equation 19.10) into Equation (1), then, yields κε A Energy = 1 CE 2 d 2 = 1 0 E 2 d 2 = 1 κε 0 AE 2 d 2 2 2 d (2) From Equation (2), we obtain expressions for the energy stored in each type of capacitor: 2 Energy air = 1 κ air ε 0 AEair d 2 and 2 Energy neo = 1 κ neo ε 0 AE neo d 2 (3) Taking the ratio of Equations (3) eliminates the unknown plate area A and plate separation d, leaving 1046 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL Energy neo Energyair = 1κ 2 neo 2 ε 0 A Eneo d 1κ 2 air 2 ε 0 A Eair d = 2 κ neo Eneo 2 κ air Eair (4) Solving Equation (4) for the energy in the neoprene-rubber-filled capacitor, we find that Energy neo ( ( κ E 2 ( 6.7 ) 1.2 ×107 V/m = neo neo Energy air = 2 κ air Eair (1.0 ) 3.0 × 106 V/m ) ( 0.075 J ) = 8.0 J 2 ) 2 57. SSM REASONING The charge that resides on the outer surface of the cell membrane is q = CV , according to Equation 19.8. Before we can use this expression, however, we must first determine the capacitance of the membrane. If we assume that the cell membrane behaves like a parallel plate capacitor filled with a dielectric, Equation 19.10 ( C = κ ε 0 A / d ) applies as well. SOLUTION The capacitance of the cell membrane is C= κε 0 A d = (5.0)(8.85 × 10 –12 F/m)(5.0 ×10 –9 m 2 ) = 2.2 × 10 –11 F –8 1.0 × 10 m a. The charge on the outer surface of the membrane is, therefore, q = CV = (2.2 × 10 –11 F)(60.0 × 10 –3 V)= 1.3 × 10 –12 C b. If the charge in part (a) is due to K + ions with charge +e (e = 1.6 × 10−19 C), the number of ions present on the outer surface of the membrane is Number of 1.3 × 10− 12 C = = 8.1×106 − 19 K + ions 1.6 ×10 C 58. REASONING Equation 19.1 indicates that the work done by the electric force as the particle moves from point A to point B is WAB = EPEA – EPEB. For motion through a distance s along the line of action of a constant force of magnitude F, the work is given by Equation 6.1 as either +Fs (if the force and the displacement have the same direction) or –Fs (if the force and the displacement have opposite directions). Here, EPEA – EPEB is given to be positive, so we can conclude that the work is WAB = +Fs and that the force points in the direction of the motion from point A to point B. The electric field is given by Equation 18.2 as E = F/q0, where q0 is the charge...
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