# Insertion of dielectric inside capacitor when capacitor is connected to a battery

When a capacitor is connected to an external voltage and a dielectric is inserted slowly across the plates, the energy of the capacitor increases. The energy preciously is $$CV^2/2$$ and after the insertion, it becomes $$KCV^2/2$$. So it increases by a factor of $$K$$.

My question is that the capacitor is connected to the battery so the potential difference across the plates has to be the same and the electric field inside the capacitor has to be the same. We were told that a capacitor stores energy in the form of an electric field (energy density = $$\epsilon E^2/2$$) so if the electric field is constant inside the capacitor how come energy is increasing by a factor of $$K$$?

• some electric field stays the same, the hypothetical one between the plates without the dielectric, and some electric field changes after the insertion of the dielectric from a hypothetical one that existed only before the insertion, to a field presently existing after the insertion. Commented Apr 30 at 12:09
• Because $\epsilon = K \epsilon_0$. Commented Apr 30 at 18:59

The energy density is $$u=\tfrac 12 \epsilon_0 \kappa E^2$$ in which $$\kappa$$ is the dielectric constant (relative permittivity). [For a vacuum $$\kappa=1$$.]

Note that we can write the formula as $$u=\tfrac 12 \epsilon_0 E^2+(\kappa-1)\tfrac 12 \epsilon_0 E^2.$$ The first term on the right can be thought of as the energy density due to the field itself; the second term as the energy stored in the dielectric by virtue of its being polarised; think of the work needed to separate positive and negative charges.

• No, I meant if Energy is stored in the form of an Electric field and if the Electric field is constant then why is energy increasing? Commented Apr 30 at 9:39
• I've added a bit to my answer, hoping that this will help with your understanding. Commented Apr 30 at 16:07

This is indeed a confusing point. The root of the issue is that there are two different definitions for the stored energy. The first definition, which is usually used in vacuum, is $$W=\frac{1}{2}\varepsilon_0 \int E^2 \mathrm{d}\mathbf{r}$$ and the second definition, which is usually used in the presence of dielectric, is $$W = \frac{1}{2} \int (\mathbf{D}\cdot\mathbf{E}) \mathrm{d}\mathbf{r} =\frac{1}{2}\varepsilon_0 \int E^2 \mathrm{d}\mathbf{r} + \frac{1}{2} \int (\mathbf{P}\cdot\mathbf{E}) \mathrm{d}\mathbf{r}.$$

They are not equivalent in general. The first equation is the total energy required to assemble the entire charge configuration, free and bound, starting from vacuum. The latter equation is the total energy required to assemble just the free charge configuration, starting with the unpolarized linear dielectric already present. The second equation is what we more commonly mean by the "energy" as we normally only control the free charge while letting the dielectric respond as it pleases.

The key point to understanding the difference is that once the static charge configuration has been set up, we have a net charge distribution which is the sum of the free and bound charge. The $$\mathbf{E}$$ is the same as the $$\mathbf{E}$$ produced by this net charge distribution alone, as if it were in vacuum. The rest of the dielectric can be removed without changing it. In other words, $$\mathbf{E}$$ only knows about the net charge distribution and does not distinguish between free and bound charge. Therefore, we can see that the first definition doesn't include any information about whether dielectric is present. It gives the same result as long as the net charge distribution is identical. You can remove all dielectric and replace all the bound charge (that was previously there) with identical free charge and the result will be the same. So the additional term in the second definition is precisely the additional energy needed to polarize the dielectric.

Here is a concrete example. Consider two geometrically identical parallel-plate capacitors, one without dielectric and the other with dielectric, charged to the same voltage. The electric field $$\mathbf{E}$$ and the charge density $$\sigma$$ are clearly identical in both cases. What's different is the free charge density $$\sigma_f$$. The first capacitor has $$\sigma_f = \sigma$$ while the second capacitor has $$\sigma_f = \varepsilon_r\sigma$$. When we normally speak of the charge in a capacitor, we are talking about $$\sigma_f$$ and not $$\sigma$$. The obvious conclusion is that the capacitance and stored energy in the second capacitor is greater. If we used the first definition, we would have concluded that the stored energy was the same. Therefore, it is clear that the second definition is what we usually refer to as the energy. Moreover, it reduces to the first one in vacuum.

This is covered in section 4.4.3 of Griffiths' Introduction to Electrodynamics.

You have already answered the question yourself by writing the electrostatic field energy density formula. The energy density $$w$$ of the electric field in a linear dielectric with absolute dielectric constant (permittivity) $$\epsilon=\epsilon_0\epsilon_r$$ is given by $$w=\epsilon E^2/2=\epsilon_0\epsilon_rE^2/2$$ where $$\epsilon_r$$ is the relative dielectric constant, you denoted by $$K$$, $$\epsilon_0$$ is the vacuum permittivity. As you said, in the capacitor connected to a battery with voltage $$V$$, the electric field doesn't change when you insert the dielectric. But $$\epsilon$$ changes from $$\epsilon_0$$ to $$\epsilon_0 \epsilon_r$$ increasing the energy density of the electrostatic field and thus also the total electrostatic energy of the capacitor by the factor $$\epsilon_r=K$$.