SM_chapter26

# SM_chapter26 - 26 Capacitance and Dielectrics CHAPTER...

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 26

SM_chapter26 - 26 Capacitance and Dielectrics CHAPTER...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online