Solution to Capacitor Problem: q=12uC, V=4V

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In summary, the question is asking for the final charge and potential difference on a 3uF capacitor connected in parallel with an uncharged 6uF capacitor. The initial situation is the 3uF capacitor charged to 12V, and the final situation is with the total capacitance being the sum of the two capacitors. The final potential difference is used to calculate the final charge on the 3uF capacitor. The common potential across both capacitors is given by the equation << exact answer deleted by berkeman >>, and the first part of the question indicates that the capacitor was initially charged by connecting it to a battery, which was then removed.
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Homework Statement


The potential across a 3uF capacitor is 12V when it is not connected to anything. It is then conntected in parallel with an uncharged 6uF capacitor. At equilibrium, the charge q on the 3uF capacitor and the potential difference V across it are
a)q=12uC, V=4V b)q=24uC, V=8V
c)q=36uC, V=12V d)q=12uC, V=6V


Homework Equations


q=CV
Parallel connections
C(total)=C1+C2


The Attempt at a Solution


The answer is a) but I am not sure why. The first part of the question confuses me. I don't really know what it means when they say that it is not connected to anything. Thanks.
 
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  • #2
Write the equation Q=CV for the initial situation, and then for the final situation. The initial situation is just the 3uF cap charged to 12V. The final situation is as you indicate, with the total capacitance the sum of the two caps. What is the final voltage across the two caps? Use that final voltage, which is across both the 3uF and 6uF capacitors now, to calculate the final charge that is on the 3uF cap.
 
  • #3
The common potential across both the capacitors after connection is given by
<< exact answer deleted by berkeman >>, i.e. total charge upon total capacitance. And for the first part I assume it means that it was charged by connecting to a battery, after which the battery is removed thereby making the charge accumulated on the plates of the capacitor constant.
 
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  • #4
shramana said:
The common potential across both the capacitors after connection is given by
<< exact answer deleted by berkeman >>, i.e. total charge upon total capacitance. And for the first part I assume it means that it was charged by connecting to a battery, after which the battery is removed thereby making the charge accumulated on the plates of the battery constant.

shramana, please do not post exact answers to homework questions. Our task here is to provide tutorial help, not the answers. That is stated clearly in the PF rules.
 
  • #5
Thank you guys!
 

1. What is the charge (q) on the capacitor?

The charge on the capacitor is 12 microcoulombs (uC).

2. What is the potential difference (V) across the capacitor?

The potential difference across the capacitor is 4 volts (V).

3. How is the charge on the capacitor related to the potential difference?

The charge on the capacitor is directly proportional to the potential difference across it. This relationship is described by the equation q=CV, where C is the capacitance of the capacitor.

4. What is the capacitance (C) of the capacitor?

The capacitance of the capacitor can be calculated using the equation C=q/V. Therefore, in this problem, the capacitance is 3 microfarads (uF).

5. How can this capacitor problem be solved?

This capacitor problem can be solved by using the equations q=CV and V=q/C. Plug in the given values of q and V to find the capacitance, then use that value to solve for any other unknowns in the problem.

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