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Homework Statement
The problem is: A 10.0-μF capacitor is charged to 15.0 V. It is next connected in series with an uncharged 5.00-μF capacitor. The series combination is finally connected across a 50.0-V battery as diagrammed in Figure P26.63. Find the new potential difference across the 5.00-μF capacitor after the switch is thrown closed.
C1= 5mF and V=unknown
C2= 10mF and charged to 15V
Homework Equations
1/Ceq= 1/C1 + 1/C2
Qtot= Ceq*V
V=Q/C
Vtotal for series= V1 + V2
The Attempt at a Solution
First I found the equivalent capacitance using the first equation and found that Ceq= 3.33mF
Then I found the charge, Q, across the capacitors using the second equation and the equivalent capacitance and found that Q=1.6E-4 C.
Then I used the last equation to try to find the potential difference across C1 using 5mF as the value for the capacitance and 1.6E-4 as the charge.
My answer is wrong and I know I need to use the potential of the 10mF capacitor somewhere but I'm not sure how? Also, I'm not sure how to incorporate the switch into the problem either?
I have tried a variety of different methods and just cannot get the right answer, I'm very stuck at this point.
Thanks