SM_chapter26

SM_chapter26 - 26 Capacitance and Dielectrics CHAPTER...

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26 Capacitance and Dielectrics CHAPTER OUTLINE 26.1 Def nition oF Capacitance 26.2 Calculating Capacitance 26.3 Combinations oF Capacitors 26.4 Energy Stored in a Charged Capacitor 26.5 Capacitors with Dielectrics 26.6 Electric Dipole in an Electric ±ield 26.7 An Atomic Description oF Dielectrics ANSWERS TO QUESTIONS *Q26.1 (a) False. (b) True. In Q = C V , the capacitance is the proportionality constant relating the variables Q and V . Q26.2 Seventeen combinations: Individual CCC 123 ,, Parallel C C CC CC CC C 12 13 23 ++ + + + ,,, Series-Parallel 11 1 3 CC C + + , 1 2 C + + , 1 1 C + + 3 1 CC C + + , 13 2 1 + + , 231 1 + + Series 111 1 , 1 + , 1 + , 1 + *Q26.3 Volume is proportional to radius cubed. Increasing the radius by a factor of 3 1/3 will triple the volume. Capacitance is proportional to radius, so it increases by a factor of 3 1/3 . Answer (d). *Q26.4 Let C 2 = NC 1 be the capacitance of the large capacitor and C 1 that of the small one. The equivalent capacitance is C eq = C CN C NN C N N C eq = + = + = + 1 1 1 1 // ( ) / This is slightly less than C 1 , answer (d). *Q26.5 We ± nd the capacitance, voltage, charge, and energy for each capacitor. (a) C = 20 µ F V = 4 V Q = C V = 80 C U = (1/2) Q V = 160 J (b) C = 30 F V = Q / C = 3 V Q = 90 C U = 135 J (c) C = Q / V = 40 F V = 2 V Q = 80 C U = 80 J (d) C = 10 F V = (2 U / C ) 1/2 = 5 V Q = 50 C U = 125 J (e) C = 2 U / V 2 = 5 F V = 10 V Q = 50 C U = 250 J (f) C = Q 2 /2 U = 20 F V = 6 V Q = 120 C U = 360 J Then (i) the ranking by capacitance is c > b > a = f > d > e. (ii) The ranking by voltage V is e > f > d > a > b > c. (iii) The ranking by charge Q is f > b > a = c > d = e. (iv) The ranking by energy U is f > e > a > b > d > c. 75
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76 Chapter 26 Q26.6 A capacitor stores energy in the electric f eld between the plates. This is most easily seen when using a “dissectible” capacitor. IF the capacitor is charged, careFully pull it apart into its compo- nent pieces. One will f nd that very little residual charge remains on each plate. When reassem- bled, the capacitor is suddenly “recharged”—by induction—due to the electric f eld set up and “stored” in the dielectric. This proves to be an instructive classroom demonstration, especially when you ask a student to reconstruct the capacitor without supplying him/her with any rubber gloves or other insulating material. (OF course, this is after they sign a liability waiver.) *Q26.7 (i) According to Q = C V , the answer is (b). (ii) ±rom U = (1/2) C V 2 , the answer is (a). *Q26.8 The charge stays constant as C is cut in halF, so U = Q 2 /2 C doubles: answer (b). *Q26.9 (i) Answer (b). (ii) Answer (c). (iii) Answer (c). (iv) Answer (a). (v) Answer (a). *Q26.10 (i) Answer (b). (ii) Answer (b). (iii) Answer (b). (iv) Answer (c). (v) Answer (b).
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SM_chapter26 - 26 Capacitance and Dielectrics CHAPTER...

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