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Say we have a capacitor of two parallel plates, full of air in between, connected to a battery that allows it to charge up. Now we deposit a dielectric plate between both capacitor’s plates, while being still connected to the battery.

My question is what will happen to the energy of the capacitor?

By doing some research, I know that the charge $Q$ increases to $kQ$ if $k$ is the dielectric constant, likewise with the capacitance $C$, which implies by $$C=\frac{Q}{V}$$ that $V$ the voltage will decrease. But about the energy I’m not so sure about it since the potential energy is: $$U=\frac{Q^2}{2C}$$ which numerator and denominator increase, but my guess is that since the charge is squared it will increase “faster” , implying that the energy will increase as well. Is that correct?

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  • $\begingroup$ If it's connected to the battery the voltage doesn't decrease. $\endgroup$
    – nasu
    Feb 20, 2023 at 14:06

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As per my knowledge(I checked my notes preparing for class 12 physics exam) If a capacitor is connected to a battery and a dielectric of dielectric constant k is inserted in capacitor then:

  • capacitance becomes k times and same for charge(as u said) but, potential V will remain same as: Q=CV (both the k will be cancelled out)
  • The potential energy will become k times U=Q^2/2C when putting value(Q = Kq let q was original charge,and C -> kC) you will get: U(final) = k * U(initial)

Finally, potential energy becomes k times

I don't know how you write the formula like you wrote but hope you understand :)

Thank you!

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