Capacitor charged then Connected to another Capacitor.

  • Thread starter Thread starter Jonnyto
  • Start date Start date
  • Tags Tags
    Capacitor Charged
AI Thread Summary
When a charged capacitor is connected to an uncharged capacitor in parallel, the total energy stored in the system decreases due to charge redistribution. Initially, the charged capacitor holds 4.0 J, but after connecting the second capacitor, the energy is halved, resulting in 2.0 J stored across both capacitors. This energy loss occurs because some energy is dissipated as heat due to the resistance and inductance in the connecting wires. The electric field in each capacitor is also reduced, leading to a decrease in energy density. Understanding these principles is crucial for analyzing energy conservation in capacitor systems.
Jonnyto
Messages
12
Reaction score
0

Homework Statement


One capacitor is charged until its stored energy is 4.0 J, the charging battery then being removed. A second uncharged capacitor is then connected to it in parallel. (a) If the charge distributes equally, what is now the total energy stored in the electric fields?(b) Where did the excess energy go?


Homework Equations


u=1/2*ε*E2 Where E is the electric field and u is the energy density(U/Volume)


The Attempt at a Solution


Okay so I'm somewhat lost. Since the charge is spread equally, and they are in parallel, so I assume they have the same capacitance. So for that I assumed that the total energy stored in each is half of that? However my reasoning is not correct since energy disappears? How can I approach the problem correctly?
 
Physics news on Phys.org
Jonnyto said:

Homework Statement


One capacitor is charged until its stored energy is 4.0 J, the charging battery then being removed. A second uncharged capacitor is then connected to it in parallel. (a) If the charge distributes equally, what is now the total energy stored in the electric fields?(b) Where did the excess energy go?


Homework Equations


u=1/2*ε*E2 Where E is the electric field and u is the energy density(U/Volume)


The Attempt at a Solution


Okay so I'm somewhat lost. Since the charge is spread equally, and they are in parallel, so I assume they have the same capacitance. So for that I assumed that the total energy stored in each is half of that? However my reasoning is not correct since energy disappears? How can I approach the problem correctly?

The charge will redistribute between the two capacitors. Each of two capacitors will have half the charge of the original capacitor. How will that affect the electric field in each capacitor?
 
Would that cut the electric field in half?
 
Jonnyto said:
Would that cut the electric field in half?

Yes, it would. What will that do to the total stored energy?
 
1/4 the original energy on each.
 
Jonnyto said:
1/4 the original energy on each.

Right again. So now you've got half the energy you started with. Where did it go?
 
Hmm that's what I'm unsure of now. Was this energy lost pushing the charge to the other capacitor?
 
Jonnyto said:
Hmm that's what I'm unsure of now. Was this energy lost pushing the charge to the other capacitor?

Well, it has to flow through a wire. What kind of property of wire might make that not free?
 
Jonnyto said:
Hmm that's what I'm unsure of now. Was this energy lost pushing the charge to the other capacitor?
Yes, but how exactly?
In the real world, the connection will have some inductance and some resistance, however small. What will that lead to?
 
Back
Top