# Charged capacitors connected together, find the final potential difference

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!

#### Attachments

• 15.9 KB Views: 624

Related Introductory Physics Homework Help News on Phys.org
gneill
Mentor

## 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
Homework Helper
Gold Member
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?

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

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
Mentor
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.