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Halliday ♦ Resnick ♦Walker FUNDAMENTALS OF PHYSICS SIXTH EDITION Selected Solutions Chapter 26 26.15 26.19 26.31 26.45 15. Let x be the separation of the plates in the lower capacitor. Then the plate separation in the upper capacitor is a − b − x. The capacitance of the lower capacitor is C = ε0 A/x and the capacitance of the upper capacitor is Cu = ε0 A/(a − b − x), where A is the plate area. Since the two capacitors are in series, the equivalent capacitance is determined from 1 a−b−x a−b 1 1 x + = . = + = Ceq C Cu ε0 A ε0 A ε0 A Thus, the equivalent capacitance is given by Ceq = ε0 A/(a − b) and is independent of x. 19. (a) After the switches are closed, the potential diﬀerences across the capacitors are the same and the two capacitors are in parallel. The potential diﬀerence from a to b is given by Vab = Q/Ceq , where Q is the net charge on the combination and Ceq is the equivalent capacitance. The equivalent capacitance is Ceq = C1 + C2 = 4.0 × 10−6 F. The total charge on the combination is the net charge on either pair of connected plates. The charge on capacitor 1 is q1 = C1 V = (1.0 × 10−6 F)(100 V) = 1.0 × 10−4 C and the charge on capacitor 2 is q2 = C2 V = (3.0 × 10−6 F)(100 V) = 3.0 × 10−4 C , so the net charge on the combination is 3.0 × 10−4 C − 1.0 × 10−4 C = 2.0 × 10−4 C. The potential diﬀerence is 2.0 × 10−4 C Vab = = 50 V . 4.0 × 10−6 F (b) The charge on capacitor 1 is now q1 = C1 Vab = (1.0 × 10−6 F)(50 V) = 5.0 × 10−5 C. (c) The charge on capacitor 2 is now q2 = C2 Vab = (3.0 × 10−6 F)(50 V) = 1.5 × 10−4 C. 31. We ﬁrst need to ﬁnd an expression for the energy stored in a cylinder of radius R and length L, whose surface lies between the inner and outer cylinders of the capacitor (a < R < b). The energy density at any point is given by u = 12 ε0 E 2 , where E is the magnitude of the electric ﬁeld at that point. If q is the charge on the surface of the inner cylinder, then the magnitude of the electric ﬁeld at a point a distance r from the cylinder axis is given by q E= 2πε0 Lr (see Eq. 26-12), and the energy density at that point is given by u= 1 q2 . ε0 E 2 = 2 2 8π ε0 L2 r2 The energy in the cylinder is the volume integral UR = u dV . Now, dV = 2πrL dr, so UR = a R q2 q2 2πrL dr = 2 2 2 8π ε0 L r 4πε0 L a R q2 R dr = ln . r 4πε0 L a To ﬁnd an expression for the total energy stored in the capacitor, we replace R with b: Ub = q2 b ln . 4πε0 L a We want the ratio UR /Ub to be 1/2, so R 1 b ln = ln a 2 a √ 1 or, since 2 ln(b/a) = ln( b/a), ln(R/a) = ln( b/a). This means R/a = b/a or R = ab. 45. (a) According to Eq. 26-17 the capacitance of an air-ﬁlled spherical capacitor is given by C0 = 4πε0 ab . b−a When the dielectric is inserted between the plates the capacitance is greater by a factor of the dielectric constant κ. Consequently, the new capacitance is C = 4πκε0 ab . b−a (b) The charge on the positive plate is q = CV = 4πκε0 ab V . b−a (c) Let the charge on the inner conductor to be −q. Immediately adjacent to it is the induced charge q . Since the electric ﬁeld is less by a factor 1/κ than the ﬁeld when no dielectric is present, then −q + q = −q/κ. Thus, κ−1 ab q = q = 4π(κ − 1)ε0 V . κ b−a