I read in my book electrostatic energy density is $ \frac{1}{2} \epsilon_0 E^2$, where $\epsilon_0$ is permittivity of vacuum and $E$ Is electric field in that region. But I want to know whether it is equally valid in dielectric medium too or do we need to replace $\epsilon_0$ with $\epsilon_r×\epsilon_0$, where $\epsilon_r$ is dielectric constant of the medium?

  • $\begingroup$ Actually expression you have given, is of electric field density, so you have to multiply with volume for energy. $\endgroup$ Commented Dec 6, 2019 at 14:36

1 Answer 1


We have $\vec{\nabla}\phi=\vec{E}$ and $\vec{\nabla}\cdot\vec{D}=\rho$.

The energy change due to the addition of a tiny amount of charge (density) $d \rho$ is: $$d U_e=\displaystyle{\int_V \phi d \rho~dV=\int_V \phi \vec{\nabla}\cdot d\vec{D}~dV=\int_V (\vec{\nabla}\cdot(\phi d\vec{D})-d \vec{D}\cdot\vec{\nabla}\phi)~dV}$$

Then the total energy is: $\displaystyle{\int\ dU_e=\int_V\int_0^D(\vec{\nabla}\cdot(\phi d\vec{D})-d\vec{D}\cdot\vec{\nabla}\phi)~dV=\int_V\int_0^D\vec{\nabla}\cdot(\phi d\vec{D})-\int_V\int_0^Dd\vec{D}\cdot\vec{\nabla}\phi~dV}$.

The first term vanishes when we integrate over the entire space (as it does in a non-dielectric media) and the only thing left is: $\displaystyle{\int_V\int_0^Dd\vec{D}\cdot\vec{E}~dV}$. Now in a linear medium: $\vec{D}=\epsilon_0\epsilon_r\vec{E}$ so $d\vec{D}=\epsilon_0\epsilon_rd\vec{E}$. Back at the energy: $U_e=\displaystyle{\epsilon_0\epsilon_r\int_V\int_0^Ed\vec{E}\cdot\vec{E}~dV=\epsilon_0\epsilon_r\int_V\int_0^Ed\vec{E}\cdot\vec{E}~dV=\dfrac{1}{2}\epsilon_0\epsilon_r\int_V\vec{E}\cdot\vec{E}~dV}$

So indeed, you're right the density is: $\dfrac{1}{2}\epsilon_0\epsilon_rE^2$.

Note here that $\rho$ is the density of free charges.

  • $\begingroup$ Your final result for the energy of the electric field in the presence of a linear dielectric is correct. However the starting point is wrong: there is no $1/2$ factor in front of your first integral for $U_e$ in term of charge density and potential. For a correct derivation, one has to start with the expression for the increment of electrostatic energy in term of the increment of free charge and eventually integrating over the increment of the resulting $\bf D$ field from zero to its final value. If the relation between $\bf E$ and $\bf D$ is linear one gets your final formula $\endgroup$ Commented Dec 6, 2019 at 16:01
  • $\begingroup$ But, $\displaystyle{U_e=\frac{1}{2}\int_V\phi\rho~dV}$ being the energy needed to bring all charges from infinity to the place where they belong; doesn't that amount to the same thing as what you said? ie: starting with the expression for the increment of electrostatic energy in term of the increment of free charge. Or maybe you can't do that, because all charges are free? $\endgroup$
    – Syrocco
    Commented Dec 6, 2019 at 16:24
  • $\begingroup$ Nevermind, I've understood my error. It's corrected, thanks!! $\endgroup$
    – Syrocco
    Commented Dec 6, 2019 at 17:24

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