333 00145 f224u1333 6 5

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: # ' ' 1 0 1 ( 0 ' 1 0 1 ( ( # 4.00 $ F . ' ) 3 $F 6 $F * ) 3 $F 6 $F * ( ) * Qtotal # CeqV # (4.00 $ F) (210 V) # 8.40 " 10!4 C . By symmetry, each capacitor carries 4.20 " 10!4 C. The voltages are then calculated via V # Q / C . This gives Vad # Q / C3 # 140 V and Vac # Q / C6 # 70 V . Vcd # Vad ! Vac # 70 V . (b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is !1 % & 1 1 !4 0 Ceq # ' ( # 4.5 $ F . Qtotal # CeqV # (4.50 $ F)(210 V) # 9.5 " 10 C , and each ) (3.00 0 6.00) $ F (3.00 0 6.00) $ F * capacitor has the same potential difference of 105 V (again, by symmetry). (c) The only way for the sum of the positive charge on one plate of C2 and the negative charge on one plate of C1 to change is for charge to flow through the switch. That is, the quantity of charge that flows through the switch is equal to the change in Q2 ! Q1 . With the switch open, Q1 # Q2 and Q2 ! Q1 # 0. After the switch is closed, Q2 ! Q1 # 315 $ C , so 315 $ C of charge flowed through the switch. EVALUATE:! When the switch is closed the charge must redistribute to make points c and d be at the same potential. 24-18 Chapter 24 24.61. (a) IDENTIFY:! Replace the three capacitors in series by their equivalent. The charge on the equivalent capacitor equals the charge on each of the original capacitors. SET UP:! The three capacitors can be replaced by their equivalent as shown in Figure 24.61a. Figure 24.61a EXECUTE:! C3 # C1 / 2 so 1 1 1 1 4 #0 0 # and Ceq # 8.4 $ F/4 # 2.1 $ F Ceq C1 C2 C3 8.4 $ F Q # CeqV # . 2.1 $ F /. 36 V / # 76 $ C The three capacitors are in series so they each have the same charge: Q1 # Q2 # Q3 # 76 $ C EVALUATE: The equivalent capacitance for capacitors in series is smaller than each of the original capacitors. (b) IDENTIFY and SET UP:! Use U # 1 QV . We know each Q and we know that V1 0 V2 0 V3 # 36 V. 2 EXECUTE:! U # 1 QV1 0 1 Q2V2 0 1 Q3V3 21 2 2 But Q1 # Q2 # Q3 # Q so U # 1 Q (V1 0 V2 0 V3 )...
View Full Document

This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.

Ask a homework question - tutors are online