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while calculating the energy stored in a capacitor of capacitance C to charge up to a voltage $V_{0}$ we say that the work done is move a infinitesimally small amount of charge $dQ$ from the negative plate to the positive an external agent has to do $(dQ) V$ amount of work which is equal to $(\frac{Q}{C})\,dQ$ which is than stored as potential energy of the capacitor Taking the integral from $ 0\, to \,Q_{0}$ bears $$\int_{0}^{Q_{0}} \frac{Q}{C}\,dQ\, = \frac{Q_{0}^2}{2C} = \frac{1}{2} C V_{0}^2$$

While finding the energy stored in the capacitor we make the same argument again just with the capacitance being $\kappa C\,$ instead of just$\, C$ .So the energy stored would be $$E=\frac{1}{2} \kappa C V_{0}^2$$

Now suppose at some certain instant the charge on the capacitor plates is $Q$ so the charge induced at the dielectric surface is $Q_{p}=Q(1-\frac{1}{\kappa})$ thus if $dQ$ charge moves through a potential difference $\frac{Q}{\kappa C}$ while moving from the negative plate to the positive plate also a $dQ(1-\frac{1}{\kappa})$ amount of charge suffers a potential difference of $- \frac{Q}{\kappa C}$. So the total change in potential energy is $$dU=dQ \frac{Q}{\kappa C}-dQ(1-\frac{1}{\kappa })(\frac{Q}{\kappa C})= $$ $$dQ\,\frac{Q}{\kappa^2 C}$$ $$\int_{0}^{Q_{0}} \frac{Q}{\kappa^2 C}\,dq\, = \frac{Q_{0}^2}{2\kappa ^2 C}=\frac{1}{2}CV_{0}^2$$ But before it was found to be $\frac{1}{2}\kappa C V_{0}^2$

where is the problem? Is there any other kind of internal energy not being considered?


2 Answers 2


The "wrong incorporation" is because you have not assumed that the capacitor as a whole is the system that you are considering and that external work is done when charging it, moving charge from one plate to the other plate with a potential $V$ across the plates.

When the charge is moved the electric field within the capacitor changes.

You can think of that electric field within the capacitor $E$ as being made up of two opposing components, $E=E_0-\dfrac {P}{\epsilon_0}$.
$E_0$, the electric field without the dielectric present and $-\dfrac {P}{\epsilon_0}$, the electric field due to the polarization of the dielectric with $P= \epsilon_0\,\chi\, E$ for a linear dielectric.

Thus $E = (1+\chi )E_0=\kappa E_0$.

Note that when charges are moved from one plate to the other it not only change $E_0$ it also change $P$ and in terms of energy storage within the capacitor.
The important parameter is $E$, the "net" field between the plates of the capacitor.

The external work done moves charges between the two plates and as a consequence of that movement $E_0$ and $P$ change which means that there is no need to add what would be an extra term relating to the "moving" of charges within the dielectric.

The energy stored per unit volume in an electric field is $\frac 12 \kappa \epsilon_0 E^2$.

Thus, the energy stored in a capacitor of plate area $A$ and plate separation $d$ is

$\frac 12 \kappa \epsilon_0 \left(\frac Vd\right)^2\cdot(Ad)=\frac 12\frac{\kappa\epsilon_0A}{d}V^2=\frac 12 CV^2$


No, the energy that you have to do is $$\tag1\int_0^Q\mathrm dU=\int_0^QV\,\mathrm dq=\int_0^Q\frac q{\epsilon_rC}\,\mathrm dq=\frac{Q^2}{2\epsilon_rC}=\frac12\epsilon_rCV_0^2$$ You did not expend any energy to move the bound charges. The voltages already encapsulated the energy differences involved.

If you are interested in the bound charges, you should be studying $\chi_e=\epsilon_r-1$, which I do suppose you had, in $\kappa-1$, but it seems to be somewhat incorporated wrongly, or else you ought to get the correct result.

  • $\begingroup$ Yes I know what is the correct answer.But the question is where and how it is "incorporated wrongly"? $\endgroup$ Commented Jun 19, 2023 at 14:21
  • $\begingroup$ I already said. The bound charges are moved by themselves, and the effect of their motion is captured in the voltages. You do not do it a second time. $\endgroup$ Commented Jun 20, 2023 at 0:37

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