Charged capacitors connected together, find the final potential difference

[The Blind Watchmaker]

Homework Statement

Homework Equations

C = Q/V

The Attempt at a Solution

Q1 = Q2
C1*V1 = C2*V2
2*V1 = 5*V2
V2 = 2/5*V1

V1+V2 =100
7/5*V1 = 100
V1 = 71.4 V

Can someone verify my working? Thanks!

 

[gneill]

The Attempt at a Solution

Q1 = Q2 ← How do you conclude this?
C1*V1 = C2*V2
2*V1 = 5*V2
V2 = 2/5*V1

Two different valued capacitors with the same potential difference will have the same charge?

 

[kuruman]

Note that before the switches are closed, Q₁ ≠ Q₂. What is the total charge on the two bottom plates before the switches are closed? What is the total charge on the two bottom plates after the switches are closed?

 

[The Blind Watchmaker]

Two different valued capacitors with the same potential difference will have the same charge?

They are in series thus Q1=Q2. I do not get why they have the same potential difference as I assume only 1 battery is used

 

[Let'sthink]

They are in series thus Q1=Q2. I do not get why they have the same potential difference as I assume only 1 battery is used

There are two issues in your understanding. When we connect two capacitors in series using only one battery q for each is taken to be the same. Just make such a connection with two different capacitors and then argue why we take charge stored by these capacitors to be the same. But in this problem they have been charged separately to the same potential difference.

 

[gneill]

They are in series thus Q1=Q2. I do not get why they have the same potential difference as I assume only 1 battery is used

Read the problem statement carefully as to how the capacitors are prepared *before* they are connected to each other.