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The energy stored in a charged capacitor of capacitance $C$ with $V$ voltage across it is:

$$U= \frac 12 CV^2$$

Whereas the work done in charging the capacitor is:

$$W= CV^2$$

This means that $(1/2)CV^2$ amount of energy gets dissipated somewhere else.

I learnt somewhere that this is dissipated as heat. But an ideal circuit having just a capacitor with a battery (without internal resistance) provides no resistance.

So:

  • How and where does half of the work done in charging a capacitor gets dissipated?
Qmechanic
  • 201,751
  • Possible duplicates: https://physics.stackexchange.com/q/187774/2451 and links therein. – Qmechanic Oct 13 '20 at 12:08
  • @Qmechanic thank you for citing that duplicate. The answer there was quite helpful and answers both of my question (i.e., how and why)! –  Oct 13 '20 at 12:19
  • The only correct answer in the cited "duplicate" is that of @my2cts – DJohnM Oct 13 '20 at 22:23
  • The second equation in this question is incorrect... There is no missing energy to explain. – DJohnM Oct 13 '20 at 22:24

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