What is the Final Potential Across the First Capacitor After Switch Closure?

AI Thread Summary
When the switch is closed, the initially charged capacitor C1 transfers charge to the uncharged capacitor C2, leading to a redistribution of charge. The total charge before the switch closure, represented as Q', is conserved, resulting in Q1 + Q2 = Q'. Using the equation Q = Cε, the final potential across C1 can be determined. The final voltage across both capacitors, which are now at the same potential, is calculated as V = (C1ε) / (C1 + C2). This analysis confirms that the potential across C1 decreases as charge is shared with C2.
phantom85
Messages
4
Reaction score
0
A first capacitor is initially connected to potential source ?. The charged capacitor is then removed
from the source and connected to a second initially uncharged capacitor. Determine the final potential
across the first capacitor long after the switch is closed.
media%2Fc03%2Fc033c36f-9c11-489c-acaa-b30d25a43852%2FphpG6QhHA.png




Homework Statement


Q1 charged by capacitor C1

Homework Equations


Q=Cε

The Attempt at a Solution

 
Physics news on Phys.org
You're going to have to give some effort to the problem. What do you think is going to happen once the the switch is closed, in terms of changes in charges and potentials, since those are the only quantities that can vary?
 
So if it let charges at C1 when connected to voltage source to be Q' then after let charges at C1 and C2 to be Q1 and Q2. It should be Q1+Q2=Q' ?
 
Yes, from conservation of charge. If you use the equation you listed, you can substitute capacitance times potential for each charge, and then that should open up the potential on C1
 
So V(C1+C2)=C1ε , and solving for V=(C1ε)/(C1+C2) is this correct?
 
Yes, that's right. V is the same across both capacitors because they become equipotential surfaces.
 
  • Like
Likes 1 person
Alright Thank you.
 
Back
Top