Voltage and energy in capacator system

AI Thread Summary
The discussion centers on a capacitor system with two 1 farad capacitors, initially charged to 1 volt. Before closing the switch, the system energy is calculated as 1/2 joule. After closing the switch, the charge redistributes, resulting in a total stored energy of 1/4 joule, indicating a loss of 1/4 joule. Participants explore potential mechanisms for this energy loss, such as heat and electromagnetic radiation. The conversation emphasizes understanding energy dissipation in capacitor systems.
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Homework Statement


There are two capacitors, 1 farad each, connected by a switch. The initial voltage of one capacitor is 1 volt. At t0, the switch is closed.

What is the system energy before closing the switch
What is the system energy, of both capacitors after closing the switch.
What is the difference in stored energy and why.


Homework Equations



V = Q/C

E = (1/2)CV^2

The Attempt at a Solution



initial energy = (1/2) (1) (1)^2 = 1/2 J
Q = CV = (1)(1) = 1 c

After switch closes, the charge is distributed over both capacitors, so each capacitor has a charge of 1/2 c
voltage is Q/C = 1/2 / 1 = 1/2 V
Energy in each cap is (1/2) (1) (1/2)^2 = 1/8
Total stored energy is 1/2 + 1/8 = 1/4

I lost 1/4 J of energy ??
 

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barryj said:
I lost 1/4 J of energy ??
Yes, yes you did.
 
So, where did it go?
 
barryj said:
So, where did it go?

Can you think of any plausible mechanisms for removing/dissipating the energy?
 
Heat, electromagnetic radiation, ?
 
barryj said:
Heat, electromagnetic radiation, ?

Yes, good.
 
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