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?

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

1 Answer 1


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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.