If the capacitor has 100 v limit and its capacitance is 1uF ,this mean that the max. energy is 1/2CV*2=5 m J
or I can put more energy on the cap.

 

milli G ????

0.5mJ is all you can put in it...

 

milli G ????

0.5mJ is all you can put in it...

You can't even cheat with electrolytics - if you supply a current limited higher voltage just under the critical leakage point, the oxide dielectric will form thicker to withstand a higher voltage - but the effectively greater spacing between the plates results in proportionately lower capacitance.

 

If this is a real world capacitor don't put 100V across it, not if you want the cap to survive for the long term.

75V is better. I would limit it to 50V.

 

If this is a real world capacitor don't put 100V across it, not if you want the cap to survive for the long term.

75V is better. I would limit it to 50V.

If its an electrolytic; the applied voltage is what keeps it formed, at a lower voltage the electrolyte very slowly etches away the oxide dielectric - if you then suddenly apply the full rated voltage without first re-forming the capacitor, it'll probably break down.

As a result of the oxide dielectric getting thinner, the capacitance increases.

If the TS is going to the trouble of calculating the energy for a given voltage/capacitance, the exact value of capacitance may be important.

 

Without parentheses, it is easy to misinterpret your equation. I am assuming that you mean \(\frac{1}{2}CV^{2}\)

I get \(5 mJ\)

My work is shown below...

\(10^{-6}*10^{4}*\ .5=.005J\ or\ 5mJ\)

 

...
\(10^{-6}*10^{4}*\ .5=.005J\ or\ 5mJ\)

Yep...