Potential difference in capacitor help

AI Thread Summary
The discussion focuses on a homework problem involving capacitors C1, C2, and C3, where the switch is initially in position A, leaving C2 and C3 uncharged. When the switch is flipped to position B, the charge and potential differences across each capacitor need to be calculated. The participants discuss the equivalent capacitance of C2 and C3 in series and how the charge from C1 will redistribute among all capacitors. They emphasize the importance of understanding that the total charge remains constant when the switch is flipped, allowing for the calculation of new potential differences. The conversation highlights the need to apply the formula V = Q/C to find the potential differences after the switch is changed.
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Homework Statement



Initially, the switch in the figure is in position A and capacitors C_2 and C_3 are uncharged. Then the switch is flipped to position B. (Figure attached)
a) Afterward, what is the charge on C_1 capacitor?
b) Afterward, what is the potential difference across C_1 capacitor?
c) Afterward, what is the charge on C_2 capacitor?
d) Afterward, what is the potential difference across C_2 capacitor?
e) Afterward, what is the charge on C_3 capacitor?
f) Afterward, what is the potential difference across C_3 capacitor?

Homework Equations



V= IR

The Attempt at a Solution



How do I start this problem?
 
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rayhan619 said:
... (Figure attached)
... How do I start this problem?

This would be a start.
 


and here is the figure...
 

Attachments

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    1.jpg
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So what do you think is going to happen?

What is the equivalent capacitance of the additional capacitors that are not initially connected?
 


we know C = Q/V
V = 100 V
C1 = 15*10^-6 F
so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C

C2 and C3 is in series. so 1/C = (1/C2 + 1/C3) = (1/30*10^-6)+(1/30*10^-6) = 66666.67
so C = 1.5*10^-5

would the Q be same after switching to point b? If yes, then we can figure out V using above equation.
 


rayhan619 said:
we know C = Q/V
V = 100 V
C1 = 15*10^-6 F
so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C

C2 and C3 is in series. so 1/C = (1/C2 + 1/C3) = (1/30*10^-6)+(1/30*10^-6) = 66666.67
so C = 1.5*10^-5

would the Q be same after switching to point b? If yes, then we can figure out V using above equation.

When the connection is broken then there is a fixed charge on the C1. When the switch connects to C2 and C3 then the charge will be shared according to what the equivalent capacitance for all 3 of them.

C2 and C3 = 2*C1

C2 and C3 equivalent C = C1

When you put them in || you have then C1 + C1 = 2 C1

Q = V*C

So with the new C = 2*old C and charge the same, then Vnew = 1/2*Vold
 


there are like 6 parts of this problem.
so for part a)
C = Q/V
V = 100 V
C1 = 15*10^-6 F
so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C
and how do I get the potential difference?
 


When uncharged capacitors are connected to the charged capacitors the common potential difference is given by V = (C1V1 + o)/(C1 + C2)
 
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