Charging Capacitors: Find Change in Charge

In summary, the switches in the three circuits will close and the charge on the left-hand capacitor will increase, decrease, or remain the same depending on the circuit.
  • #1
downwithsocks
37
0

Homework Statement


The figure shows three circuits, each consisting of a switch and two capacitors, initially charged as indicated (top plate positive). After the switches have been closed, in which circuit (if any) will the charge on the left-hand capacitor (a) increase, (b) decrease, and (c) remain the same?

[PLAIN]http://img714.imageshack.us/img714/3369/capacitance.png


Homework Equations


q = CV


The Attempt at a Solution


I really am not sure how to approach the problem. I know I should find the voltages before and after the switch is closed in each situation, but I'm not really sure how to do that with the "q"s in there.
 
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  • #2
charges will move until the voltage over each of the 2 capacitors will be equal,
i.e. V will be constant over all the circuit and that means that there is no electric field so no more charges can be moved.

V=q/C
grad(V)=E
 
  • #3
You only need to know what direction the current will flow when the switch is closed. You're given charge and capacitance for each capacitor (albeit symbolically). You should be able work out the relative sizes of the voltages on each cap.
 
  • #4
Are 6q and 3q capacitances then? I feel really stupid that I just can't work out how to solve this.
 
  • #5
downwithsocks said:
Are 6q and 3q capacitances then? I feel really stupid that I just can't work out how to solve this.

6q and 3q are charges on the capacitors of value 3C and C. You don't need to know what the actual values of q (coulombs) and C (farads) are.
 
  • #6
So then the first one won't change, the second one will increase, and the third one will decrease?
 
  • #7
It is so.
 
  • #8
Thanks so much, I was confused because I thought the C values were charges in coulombs, so I didn't know how to use my knowledge of voltage given only charges.
 

FAQ: Charging Capacitors: Find Change in Charge

What is a capacitor?

A capacitor is an electronic component that stores electrical energy in an electric field. It is made up of two conductive plates separated by an insulating material known as a dielectric.

How do you charge a capacitor?

A capacitor can be charged by connecting it to a power source, such as a battery, through a conductive material, like a wire. The positive terminal of the power source is connected to one plate of the capacitor, and the negative terminal to the other plate. This causes an accumulation of charge on the plates, creating an electric field between them.

3. What is the equation for finding the change in charge of a capacitor?

The equation for finding the change in charge of a capacitor is Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts. This equation shows the direct relationship between the capacitance and the voltage of a capacitor.

4. How does the capacitance affect the charge on a capacitor?

The capacitance of a capacitor determines how much charge it can hold at a given voltage. A larger capacitance means that the capacitor can hold more charge, while a smaller capacitance can hold less charge at the same voltage.

5. What factors can affect the change in charge of a capacitor?

The change in charge of a capacitor can be affected by the voltage, capacitance, and resistance of the circuit. Temperature can also play a role, as changes in temperature can alter the properties of the dielectric material and therefore affect the capacitance of the capacitor.

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