b) After the switch is flipped to the position B, the battery is no longer connected to the contour and the charge redistributes between the ca- pacitors as shown in Fig.2. Notice that I used the fact that the segment between the capacitors C₂ and C3 has to be neutral (therefore, they have the same charge), but the segments connecting C₁ to C₂ and C₁ to C3 are not neutral. What can you say about the sum of charges Q₁ and Q2? + 100V S -Q₁ Q₂ =Q₂ Q2 Q₂ FIG. 2: The scheme for Problem 1b

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Hi, I know there are multiple parts to this but I only need help is with part B can you help me with Part B. Thank you

Problem 1: Initially, the switch in Fig 1. is in its position A and
capacitors C₂ and C3 are uncharged. Then the switch is flipped to
position B. Afterward, what are the charge on and the potential dif-
ference across each capacitor?
Partial answer: AV₁ = 55 V, AV₂ = 33.5 V.
a) While the capacitor is in position A, as shown in Fig.1, com-
pute the charge Q accumulated on the plates of the capacitor C₁.
b) After the switch is flipped to the position B, the battery is no longer
connected to the contour and the charge redistributes between the ca-
pacitors as shown in Fig.2. Notice that I used the fact that the segment
between the capacitors C₂ and C3 has to be neutral (therefore, they have
the same charge), but the segments connecting C₁ to C₂ and C₁ to C3 are
not neutral. What can you say about the sum of charges Q₁ and Q2?
Switch
100V
B
+
C₁ =ISμF
+
100V
_C₁₂=20μF
FIG. 1: The scheme for Problem 1
-Q₁
C₂=30μF
Q₂
Q₂
Q₂
EQ2
FIG. 2: The scheme for Problem 1b
c) Use Kirchhoff's loop law to get another relation between charges Q₁ and Q₂. Starting from point B
in Fig.2, move counterclockwise along the loop and register the potential differences that you encounter
when crossing the capacitors (pay attention to the signs - when you move from a positively charged to a
negatively charged plate, the potential is decreasing). The sum of all potential differences has to be zero.
d) You answers to parts b) and c) give you a system of two equations that you can solve to find individual
values of charges Q₁ and Q₂. Solve it to find the values of the charges, and then compute the potential
differences across each capacitor.
Transcribed Image Text:Problem 1: Initially, the switch in Fig 1. is in its position A and capacitors C₂ and C3 are uncharged. Then the switch is flipped to position B. Afterward, what are the charge on and the potential dif- ference across each capacitor? Partial answer: AV₁ = 55 V, AV₂ = 33.5 V. a) While the capacitor is in position A, as shown in Fig.1, com- pute the charge Q accumulated on the plates of the capacitor C₁. b) After the switch is flipped to the position B, the battery is no longer connected to the contour and the charge redistributes between the ca- pacitors as shown in Fig.2. Notice that I used the fact that the segment between the capacitors C₂ and C3 has to be neutral (therefore, they have the same charge), but the segments connecting C₁ to C₂ and C₁ to C3 are not neutral. What can you say about the sum of charges Q₁ and Q2? Switch 100V B + C₁ =ISμF + 100V _C₁₂=20μF FIG. 1: The scheme for Problem 1 -Q₁ C₂=30μF Q₂ Q₂ Q₂ EQ2 FIG. 2: The scheme for Problem 1b c) Use Kirchhoff's loop law to get another relation between charges Q₁ and Q₂. Starting from point B in Fig.2, move counterclockwise along the loop and register the potential differences that you encounter when crossing the capacitors (pay attention to the signs - when you move from a positively charged to a negatively charged plate, the potential is decreasing). The sum of all potential differences has to be zero. d) You answers to parts b) and c) give you a system of two equations that you can solve to find individual values of charges Q₁ and Q₂. Solve it to find the values of the charges, and then compute the potential differences across each capacitor.
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