Capacitor is charged then connected to a capacitor and battery

In summary, the conversation discusses the charge on capacitor C_2 after it is connected to capacitor C_1 and a second battery in a circuit. The initial charge on C_1 is related to the charge on C_2 through the loop rule. The solution involves labeling the charges on the plates and equating the sum of the charges before and after the switching occurs.
  • #1
TheEvenfall
3
0
Hello PF,

Homework Statement


The capacitor C_1 is connected to a battery as shown. After being charged, it is connected to the capacitor C_2 and the second battery as shown. When all the charges come to rest, what will be the charge on capacitor C_2?
http://imgur.com/4ciJRrn

Homework Equations


C = Q/V

The Attempt at a Solution


The initial charge on C_1 is Q_1 = C_1 V_1
After it is connected to the other circuit and in steady state, I used the loop rule to get the equation (q_1/C_1) + (q_2/C_2) = V_2, where q_1 is the new charge on C_1 and q_2 is the charge on C_2.
Now the problem is relating q_1, q_2 and Q_1. According to my professor's solution, -Q_1 = -q_1 + q_2.
He got that equation by doing what's shown here: http://imgur.com/WvKJ6rQ, and I have no idea where he got that from, and this is the point that I'm stuck on and why I'm posting this.
Using that equation we can find q_2 easily, but I don't understand how he got it.

Thank you.
 
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  • #2
TheEvenfall said:
Hello PF,

Homework Statement


The capacitor C_1 is connected to a battery as shown. After being charged, it is connected to the capacitor C_2 and the second battery as shown. When all the charges come to rest, what will be the charge on capacitor C_2?
http://imgur.com/4ciJRrn

Homework Equations


C = Q/V


The Attempt at a Solution


The initial charge on C_1 is Q_1 = C_1 V_1
After it is connected to the other circuit and in steady state, I used the loop rule to get the equation (q_1/C_1) + (q_2/C_2) = V_2, where q_1 is the new charge on C_1 and q_2 is the charge on C_2.
Now the problem is relating q_1, q_2 and Q_1. According to my professor's solution, -Q_1 = -q_1 + q_2.
He got that equation by doing what's shown here: http://imgur.com/WvKJ6rQ, and I have no idea where he got that from, and this is the point that I'm stuck on and why I'm posting this.
Using that equation we can find q_2 easily, but I don't understand how he got it.

Thank you.

The lower plate of C1 and left plate of C2 along with the wire connecting them form an isolated "island" for charge. Whatever charge is on that island must remain, and there's no way for new charge to appear there.

So label the plates for their charges before and after the switching occurs. If the total charge on the island can't change then you can equate the sum of the charges on the plates for the before and after conditions.
 
  • #3
gneill said:
The lower plate of C1 and left plate of C2 along with the wire connecting them form an isolated "island" for charge. Whatever charge is on that island must remain, and there's no way for new charge to appear there.

So label the plates for their charges before and after the switching occurs. If the total charge on the island can't change then you can equate the sum of the charges on the plates for the before and after conditions.

Got it, thanks a lot.
 

Related to Capacitor is charged then connected to a capacitor and battery

1. What happens when a charged capacitor is connected to another capacitor and a battery?

When a charged capacitor is connected to another capacitor and a battery, the excess charge from the first capacitor will flow into the second capacitor through the battery, causing it to become charged as well. This process is known as capacitive coupling.

2. Can a capacitor be charged multiple times?

Yes, a capacitor can be charged multiple times. However, the maximum amount of charge it can hold is limited by its capacitance value.

3. How long does it take for a capacitor to charge?

The time it takes for a capacitor to charge depends on its capacitance and the amount of current flowing through it. The larger the capacitance and the higher the current, the longer it will take to charge.

4. How does the voltage across a capacitor change when it is connected to a battery?

When a capacitor is connected to a battery, the voltage across the capacitor will increase until it reaches the same voltage as the battery. This is because the battery supplies a constant voltage, but the capacitor acts as a temporary voltage source until it becomes fully charged.

5. What is the purpose of connecting a charged capacitor to another capacitor and a battery?

The purpose of connecting a charged capacitor to another capacitor and a battery is to transfer the excess charge from the first capacitor to the second capacitor, effectively increasing the overall capacitance in a circuit. This can be useful in certain electronic applications, such as filtering or signal processing.

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