When we introduce a dielectric of dielectric constant $K$ in the space between the conducting plates(slowly) of a capacitor connected to a battery with a voltage $V$ across it, we find that an extra $(K-1)\frac{CV^2}{2}$ amount of work is done by the external agent inserting the dielectric material. My question is, where does this work go? Is it used as energy to overcome the force experienced by dielectric by the conducting plates while it is being inserted, or is it used to change the arrangement of the dipoles in the dielectric to produce net bound charge, or is it (or a portion of it) lost in the form of heat?

I think it is a combination of all of these.


1 Answer 1


So alright, this question is a fun one. In the overall scheme, the system indeed gain a energy of $(K-1)\frac{CV^2}{2}$. But I guess you're asking why and how this happens.


Because for any configuration of an electrostatic (or magnetostatic, the calculations can be made analogous) system, we can define electric potential energy. There are two equivalent ways of doing so,

$$U_{total}=\frac{1}{2}\sum_{i\in\text{All particles}}\sum_{j\in\text{All other particles}} \frac{q_iq_j}{4\pi\epsilon_0|\mathbf r_i-\mathbf r_j|}$$

$$U_{total}=\int_\text{entire space} \frac{\epsilon(\mathbf r)\,\mathbf E(\mathbf r)\cdot\mathbf E(\mathbf r)}{2}\,dV=\int_\text{entire space} \frac{\mathbf D(\mathbf r)\cdot\mathbf E(\mathbf r)}{2}\,dV$$

The problem with the first (and more conventional) method is that you need to account for all charges everywhere, free or bound. In the second one, knowing the field configuration is enough, and thus it is easier to compute in this system.

In our system, the electric field exists only in the interior of the capacitor and it has a constant (independent of the dielectric constant in that vertical section) amplitude since $V=\Delta V=-\int_{y=0}^{h} \mathbf E(x,y)\cdot\mathrm d\vec y$, i.e. the potential of the capacitor is constant across different $x$ and movement across $y$ is homogeneous: $\mathbf E(x,y) = -(V/h)\,\hat y$.

So we can simply integrate this in the volume of the capacitor: $$U_{total}=\int_\text{entire space} \frac{\epsilon(\mathbf r)\,\mathbf E(\mathbf r)\cdot\mathbf E(\mathbf r)}{2}\,dV=\int_{x=0}^d \int_{y=0}^h \int_{z=0}^d \frac{K\epsilon_0 V^2}{2h^2}\,dz\,dy\,dx$$ $$=\frac{K\epsilon_0 V^2dhL}{2h^2}=K\frac{\epsilon_0dL}{2h}V^2=K\frac{C}{2}V^2$$

Since in the initial state $K=1$, difference is $(K-1)\frac{CV^2}{2}$.

Notice that I just calculated $\Delta E_{total}$ only in terms of the change of electrostatic potential energy: I assume there is no kinetic energy, and that there is no other contributions to the potential term. However there is no explanation as to how we add energy to the system. To see a more complete picture, we need to investigate


How do we introduce energy into this system? There are two plausible ways: 1) Mechanical work done on the dielectric (pushing/pulling), and 2) The battery. First way is a simple $W=Fx$, but the second one is the more important but occasionally ignored $W=\int_{t=0}^{T}IV\,dt=V\,\Delta Q$.

The battery always keep the capacitor voltage at $V$. Now ask yourself: If you increase the capacitance of a capacitor, does its charge change? The answer is yes, and this change is $\Delta Q = C_f\,V - C_i\,V$ with $C_f=K\,C_i$, so $\Delta Q = (K-1)CV$. Notice that this is positive, and thus the battery needs to do work in order to establish the new equilibrium, $W_{battery} = V\,\Delta Q = (K-1)\,CV^2$.

So we found that the battery adds twice the energy needed to increase the potential energy. The rest is either converted to kinetic energy, or we hold the system back from going in with an acceleration and thus take back (negative) work, $W_{mechanical}=-(K-1)\frac{CV^2}{2}$.

The answer to your question is, the energy is used to hold the new charges in equilibrium. However we do not push the dielectric in, on the contrary we need to hold it back from going in by itself. The part that does the work is the battery that maintains the voltage of the capacitor.


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