A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by,
New electrostatic energy can be calculated as
Loss in electrostatic enegy = E - E'
= 1.2 x 10-5 - 0.6 x 10-5
= 0.6 x 10-5
= 6 x 10-6 J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.
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8 Comment(s) on this Question
Thank you
common potential =c1c2\\c1+c2
How can we now that it is connected in series
It is parallel not series.
How can we know that the capacitors are connected in series?
How can we conclude that capacitors are in series
In this question it is not written that uncharged capacitor is connected in series then how you take in series according to the solution??
what if the unchanged capacitor is connected to parallel with it
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