When we insert dielectric in a charged capacitor then, dielectric is attracted by the capacitor which makes dielectric being "sucked up" by the capacitor. Now, in order to derive the formula for force on dielectric, we apply energy conservation which I feel to be wrong as heat will be lost and hence energy conservation should not be applied, furthermore I feel that the energy term in the derivation corresponding to work done by force is actually heat released in the process.Where am I going wrong, also see [ Heat produced when dielectric inserted in a capacitor ] for more clarification of the doubt. Thanks
Electrostatics – Force on Dielectric Inserted in Charged Capacitor
classical-electrodynamicselectrostatics
Related Solutions
Let us first consider a capacitor that is charged and not connected to a battery or other electric power source.
I think you need to take into account that the Electric field in the capacitor is reduced by inserting the dielectric and also the voltage drops between the plates by inserting the capacitor. The dielectric is pulled in to the cap as the capacitor loses electrical potential energy as its voltage is decrease.
The dielectric in the capacity gets polarized so that the surface of the dielectric facing the positively charged plate becomes negatively charged and the surface of the dielectric facing the negatively charged plate becomes positvely charged. The charges on each side of the dielectric are less than the charges on the plates. Overall the electric field inside the capacitor (inside the dielectric) is reduced because the effective total charge on each side of it is reduced.
The electric field inside the capacitor $E'$ is reduced by
$E'={1 \over k} E~~~~~$ or $~~~~~E'={1 \over \epsilon_r} E$
The reduction in electric field comes with a reduction in the potential between the two plates so the energy stored in the capacitor is reduced as the same ammount of charge is stored in it, but the voltage is now lower - so that if it is discharged the power term $VI$ would be smaller as $V$ is lower.
If the battery is still attached then as the dielectric is inserted the capacitance is increased and the capacitor charges up more. The charge is multiplied by dielectric constant, $k$, and then the battery supplies the extra charge and so it loses energy.
As the battery is attached the electric field inside the dielectric is the same as the original electic field in between the plates without the dielectric, but now the charge is higher on the plates and there is again charge induced on the surfaces of the two dielectrics. I think the upper of the two diagrams in the question is correct because the electric field in the gap is the same whether with and without the dielectric inserted.
There will be addition charge density on the plates in regions II and III, but these will be balanced by the extra opposite charges on the surface of the dielectric in II and III so that the upper diagram is correct.
A dielectric slab will be attracted towards any source of electric field, because it is essentially a collection of dipoles. Dipoles in fields align with the fields. If the field varies in space, the dipole is drawn to the local maximum of the field. An intuitive, everyday example of this is the way iron filings are drawn to a magnet.
In the case of a constant voltage capacitor, the battery supplies the extra energy. We can see this most simply in the case that we replace the battery with a very large capacitor of capacitance $C$ connected in parallel with our original capacitor with capacitance $c_0 \ll C$. Originally, each capacitor is at a voltage $V_0$, and the large capacitor has charge $Q_0=CV_0$ while the small capacitor has charge $q_0=c_0V_0$. As we insert the dielectric, $c_0$ changes by $\Delta c$. The capacitors then settle in the new condition of equal voltage
$$\frac{Q_0-\Delta q}{C} = \frac{q_0 + \Delta q}{c_0 + \Delta c}$$
Solving for $\Delta q$ using the fact that $c_0 +\Delta c \ll C$ gives
$$\Delta q = Q_0 \frac{c_0 + \Delta c}{C} - q_0$$
From this it is fairly easy to show that the energy change in the original capacitor is
$$\Delta U_c=\frac{1}{2}\Delta c V_0^2$$
What about the change in the large capacitor? There we have
$$\Delta U_C = \frac{1}{2C}((Q-\Delta q)^2 - Q^2) \approx -\frac{\Delta q\, Q}{C}$$
Substituting gives
$$\Delta U_C=-\Delta c V_0^2$$
So overall the system ends with a lower energy, and the slab should be attracted.
Best Answer
This problem is equivalent to releasing a mass which is at the end of an unextended vertical spring.
The mass loses gravitational potential energy whilst at the same time gains elastic potential energy and kinetic energy.
When the mass reaches the static equilibrium position it has kinetic energy and so overshoots that static equilibrium position to eventually stop when the loss of gravitational potential energy is equal to the gain in elastic potential energy.
At that maximum downward excursion there is a net upward force on the mass so it starts to move upwards losing elastic potential energy whilst at the same time gaining gravitational potential energy and kinetic energy.
The mass passes through the static equilibrium position and carries on until it reaches its initial position with the spring unextended and the mass at rest.
The process repeats itself the the mass oscillates about the static equilibrium position.
In the real world friction is present and so the mass undergoes damped harmonic motion eventually stopping at the static equilibrium position.
The spring-mass system now has less mechanical energy than it started with (it is actually half as much) because of the negative work done on the sysyem by the frictional forces - heat is generated.
Now with your capacitor and dielectric a very similar thing happens.
Start with the dielectric (same size as one of the plates of the capacitor) just outside the capacitor and release the dielectric.
There is a force on the dielectric which pulls it into the capacitor and electric potential energy is converted into kinetic energy of the dielectric.
When the dielectric reaches the static equilibrium position it has kinetic energy and will overshoot the static equilibrium position and eventually end up at rest jest outside the capacitor on the other side from where it started.
Although at rest there is a force on the dielectric pulling it into the capacitor and so the dielectric now starts moving in the opposite direction finishing up at rest at its starting point.
If one does not ignore dissipative process then the dielectric will undergo a motion which will result in it finally being at rest in the capacitor - the static equilibrium position.
Overall the dielectric-capacitor (and possibly battery) would have lost electrical energy and generated heat.
When you work out the force on a dielectric you can do it by ignoring dissipative processes and allow the dielectric to move from its current position a small distance .$dx$.
What you do next depends on whether the capacitor is isolated (constant charge $Q$ or connected to a battery (constant potential difference $V$).
If the capacitor has a constant charge then the change in energy stored $dU$ in the capacitor $C$ when the dielectric is moved a distance $dx$ in a direction parallel to the sides of the plates is $dU = -F\,dx$ where $F$ is the force on the dielectric.
So $F = -\frac{dU}{dx} $ with $U= \frac 1 2 \frac {Q^2}{C}$ which gives $ F = \frac 1 2 \frac {Q^2}{C^2}\frac{dC}{dx}$.
Knowing an expression for $C$ in terms of $x$ the force $F$ can be found.
Conservation of energy has been used in the absence of any dissipative processes, there are no frictional forces and no losses within the dielectric when its polarisation changes.