Recent content by 190793

Of course! I should have known that. Part B of the question asks for the potential difference across the big capacitor after the switch is closed which would be 50-23.3 Thank you so much for your help I FINALLY understand!

More specifically it is 23.34 and 11.67? I found that by trial and error, what is the correct method?

No of course that doesn't work that just brings me back to 15!

Alright so V5=17.5-V10 And V10=Q/C

I'm not sure where to go from here. Is the charge on the 5μF capacitor 1.64E-4 (the charge I found earlier using Ceq) minus the charge on the 10μF capacitor?

Ok so if I find the charge for the 10μF capacitor, can I subtract 15C from it to find the charge of the 5μF capacitor?

Ok so Ceq remains the same. Does Q? I was under the impression that the charge is equivalent for two capacitors in series.

Gosh I'm sorry that isn't making things very clear for me. Would the 5-microFarad capacitor take the bulk of the 50V received from the battery since the 10-microFarad capacitor already has an initial charge? And after the battery is turned on, is the total potential difference of the circuit 50V...

So if the 10microF capacitor had an initial charge of 15V, how can I incorporate that into finding the answer?

I didn't know how to get the μ symbol and mistakenly thought it could be substituted with an m! Thanks for the heads up!

I see! So the fact that the 10mF capacitor was previously charged to 15V makes no difference? Thank you so much for your help the question was driving me crazy!

Well since the voltage is inversely proportional to the capacitance (V=Q/C) I want to say that the 5mF will have more voltage? Twice as much? I'm not sure if this is correct though.

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...