Two capacitors are connected in parallel

AI Thread Summary
The discussion revolves around solving a problem involving two capacitors in parallel connected to a 12 V battery. The equivalent capacitance is calculated as 38.3 μF. The main challenge is determining the current flowing in the circuit, especially after the capacitors are charged, where the current approaches zero. Participants clarify that the battery maintains a constant voltage, and the current can be expressed as a function of charge over time. The importance of distance in capacitance is also briefly mentioned, linking it to the formula C=εA/d.
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Homework Statement


Two capacitors (C1 = 9.3 μF and C2 = 29 μF) are connected in parallel across a 12 V battery.
a) Find the equivalent capacitance of the two capacitors.
Ceq=C1+C2=9.3+29=38.3μF

After some time has passed (use for rest of problem):
b) How much current flows in this circuit?
____A

c) Find the charge stored on each capacitor.
Q1 =____μC
Q2 =____μC

d) Calculate the potential energy stored by each capacitor:
U1 =____μJ
U2 =____μJ



Homework Equations



C=Q/V

U=(1/2)*C*V^2

I=dq/dt

The Attempt at a Solution




part b is where i got stuck. I don't know how to find the current flowing in the circuit if the time wasnt given to me. Current is a function of (dq/dt).
I know after for very long time, the current flow into the capacitor will be zero since the battery used all of its energy to charge the capacitor, therefore no current flow.
Can someone help me out? Thank you so much
 
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The battery doesn't necessarily use up all its energy. You are in fact to assume that the battery voltage maintains 12V thruout.

Assume a small resistor in series with the battery and capacitor bank. What is the expression for current if the capacitors are initially uncharged and then the circuit is closed?

There is a formula for the energy stored in a capacitor as a function of its voltage and capacitance in your textbook, I'm sure. Or you can derive it by determining the total work done in moving charge Q from the - (bottom) plate to the + (top) plate where Q = CV. The plates start with q=0, a small amount of charge dq is removed from the bottom plate and moved to the top plate; this process is repeated until the full chatge Q sits on the top plate. As more and more charge is moved, the potential difference V increases, so the work done in moving a differential charge dq increases as the charges are moved.
.
 
Why distance is important in capacitance

We know that capacitance can be calculated by the formula C=εA/d where 'd' is the distance between parallel plates. But why distance should matter? Can anyone explain please. Thank you.
 
Do you understand the answer to your last question (post 1) before embarking on the next?
 
thanks guys... apparently the answer is current is reaching zero after sometimes.
and somehow we need to do the derivative. but i didnot do that, i just plug in the number
lucky guess
 
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