Voltage and energy in capacator system

Heat or electromagnetic radiation are both plausible mechanisms for removing or dissipating the energy.In summary, two 1 farad capacitors connected by a switch have an initial energy of 1/2 J and a charge of 1 c. When the switch is closed, the charge is distributed evenly between the two capacitors, resulting in each capacitor having a charge of 1/2 c and a voltage of 1/2 V. The total stored energy after the switch is closed is 1/4 J, resulting in a loss of 1/4 J of energy. This energy can be dissipated through mechanisms such as heat or electromagnetic radiation.
  • #1
barryj
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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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  • #2
barryj said:
I lost 1/4 J of energy ??
Yes, yes you did.
 
  • #3
So, where did it go?
 
  • #4
barryj said:
So, where did it go?

Can you think of any plausible mechanisms for removing/dissipating the energy?
 
  • #5
Heat, electromagnetic radiation, ?
 
  • #6
barryj said:
Heat, electromagnetic radiation, ?

Yes, good.
 

FAQ: Voltage and energy in capacator system

1. What is the relationship between voltage and energy in a capacitor system?

The voltage and energy in a capacitor system are directly proportional. This means that as the voltage increases, so does the energy stored in the capacitor. Similarly, as the voltage decreases, the energy stored in the capacitor decreases as well.

2. How is the energy stored in a capacitor system calculated?

The energy stored in a capacitor system can be calculated using the formula E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts.

3. Can the energy in a capacitor system be negative?

No, the energy in a capacitor system cannot be negative. This is because the energy stored in a capacitor is always positive and is released when the capacitor discharges.

4. What happens to the voltage and energy in a capacitor system when a dielectric material is inserted?

When a dielectric material is inserted into a capacitor, the voltage remains the same but the energy stored in the capacitor increases. This is because the dielectric material increases the capacitance, which in turn increases the energy stored.

5. How does the type of dielectric material used affect the energy stored in a capacitor system?

The type of dielectric material used can significantly affect the energy stored in a capacitor system. Different dielectric materials have different dielectric constants, which determine the capacitance of the capacitor. Therefore, using a dielectric material with a higher dielectric constant will result in a higher capacitance and therefore, a higher energy stored in the capacitor system.

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