How Does Connecting Capacitors in Parallel Affect Charge and Potential?

[Jalo]

Homework Statement

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Consider a capacitator of capacity C1 charged with a charge Q. Suppose you connect it in parallel to another capacitator, C2, initially uncharged.

Find the final charge and difference of potencial at each condensator.

Homework Equations

The Attempt at a Solution

I know that the initial charge Q will be equal to:

Q = q1+q2 , where q1 and q2 is the final charge of capacitator 1 and 2.
The difference of potential, V, is equal in both the capacitators. Therefore I can say that

q1 = Q - q2 = Q - C2/V = [ QV - C2 ] / V

I don't know where to go from here. I don't think I can get anywhere... If anyone could point me in the right direction I'd really appreciate!

Thanks.

 

[aralbrec]

You're thinking in the right direction.

Initially there is a charge on C1 equal to Qi1. There is no charge on C2.

In the end, the voltage across both capacitors is equal, so:

Vf1 = Vf2; Qf1/C1 = Qf2/C2

I'm using Qi as initial charge and Qf as final charge.

Between the initial state and the final state, charge moves from capacitor C1 to capacitor C2. If you assume a charge dQ moves from C1 to C2, you should be able to say something about Qf1 and Qf2 in terms of Qi1.

 

[ehild]

Homework Statement

;-------| |---------;
| |
| |
'-------| |---------'

Consider a capacitator of capacity C1 charged with a charge Q. Suppose you connect it in parallel to another capacitator, C2, initially uncharged.

Find the final charge and difference of potencial at each condensator.

Homework Equations

The Attempt at a Solution

I know that the initial charge Q will be equal to:

Q = q1+q2 , where q1 and q2 is the final charge of capacitator 1 and 2.
The difference of potential, V, is equal in both the capacitators. Therefore I can say that

q1 = Q - q2 = Q - C2/V = [ QV - C2 ] / V

I don't know where to go from here. I don't think I can get anywhere... If anyone could point me in the right direction I'd really appreciate!

Thanks.

Q=CV. AS the potential difference is the same across the parallel connected capacitors, q₁=C₁V and q₂ = C₂V. q₁+q₂=Q. Can you find V in terms of Q and C1, C2? ehild

 

[Jalo]

Q=CV. AS the potential difference is the same across the parallel connected capacitors, q₁=C₁V and q₂ = C₂V. q₁+q₂=Q. Can you find V in terms of Q and C1, C2?


ehild

Oh, so simple... I got to the correct result. Thank you very much!

 

[Nickie]

To find the final charge and difference of potential at each capacitator, you can use the equation for capacitance in parallel:

C = C1 + C2

Since the capacitators are connected in parallel, the voltage across them will be the same. This means that the final charge on both capacitators will be equal, let's call it Qf. So we can write:

Qf = C1V = C2V

And since C1 = Q/C1 and C2 = Qf/C2, we can rewrite the equation as:

Q/C1 = Qf/C2

Solving for Qf, we get:

Qf = C2Q/C1

Now, using the equation for capacitance in parallel, we can substitute C2 for C1 in the above equation:

Qf = (C1 + C2)Q/C1

Simplifying, we get:

Qf = Q + C2Q/C1

So the final charge on the first capacitator will be Q + C2Q/C1, and the final charge on the second capacitator will be C2Q/C1. To find the difference of potential at each capacitator, we can use the equation V = Q/C, where C is the capacitance of each capacitator. So the difference of potential at the first capacitator will be:

V1 = (Q + C2Q/C1)/C1 = Q/C1 + C2Q/C1^2

Similarly, the difference of potential at the second capacitator will be:

V2 = (C2Q/C1)/C2 = Q/C1

So the final charge and difference of potential at each capacitator will be:

Final charge at C1 = Q + C2Q/C1
Final charge at C2 = C2Q/C1
Difference of potential at C1 = Q/C1 + C2Q/C1^2
Difference of potential at C2 = Q/C1