Variable capacitor energy vs setting

Thread Starter

Ylli

Joined Nov 13, 2015
788
Reading the thread on "Exotic Propulsion", this came to mind: It's a variation on the putting two caps with different charges in parallel. I'm sure I'm missing something simple here, but I've got a brain block.

Let's start with a 20 - 365 pF variable capacitor (no leakage) and a 100 volt battery:

Set the capacitor to 20 pF and connect it to the battery.
Q = C*V = 20p*100 = 2000 pC
W = 0.5*C*V² = 0.5*20p*100² = 100,000 pJ

Now disconnect the battery and set the variable cap to 365 pF
V = Q/C = 2000pC/365p = 5.48 volts
W = 0.5*C*V² = 0.5*365p*5.48² = 5481 pJ

Where did the energy go? Did it radiate into space?


Now lets do it the other way around. Charge the 365 pF cap to 100 volts:

Set the capacitor to 365pF and connect it to the battery.
Q = C*V = 365p*100 = 36,500 pC
W = 0.5*C*V² = 0.5*365p*100² = 1,825,000 pJ

Now disconnect the battery and reset the cap to 20pF
V = Q/C = 36,500p/20p = 1825 volts
W = 0.5*C*V² = 0.5*20p*1825² = 33,306,250 pJ

Where did the energy come from?
 

MrChips

Joined Oct 2, 2009
19,379
Simple. Mechanical energy.

Take the second case. In order to reduce the capacitance you would have had to do something such as pull the plates apart. There is an electric field between the two plates with positive charges on one plate and negative charges on the other. You have to expend energy to pull the plates away from each other.

The reverse happens in the first case. Energy is released as the plates are brought closer to each other.
 

Thread Starter

Ylli

Joined Nov 13, 2015
788
Simple. Mechanical energy.

Take the second case. In order to reduce the capacitance you would have had to do something such as pull the plates apart. There is an electric field between the two plates with positive charges on one plate and negative charges on the other. You have to expend energy to pull the plates away from each other.

The reverse happens in the first case. Energy is released as the plates are brought closer to each other.
Duh. I though about the mechanical effort required to move the shaft on the capacitor, but was thinking in terms of the friction, and knew that couldn't be it. But yes, a force against the charge differential makes sense.
 
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