Why is no heat produced when dielectric inserted slowly in charged capacitor When a dielectric is inserted slowly in a charged capacitor (connected to battery), then it has been told to me that no heat is evolved which appears unreasonable to me. I think it in this way that when dielectric is inserted slowly then in an interval of time it just gets "sucked up" for a very small distance as compared to when moving fast, so the extra charge by the battery will be supplied in a very short range compared to when fast moving, but the speed of the charge must have been same and finally, all will come to rest establishing an equal charge concentration on the plates in both the cases. So things might take a little longer when inserted slowly but the amount of heat released should be same. Where am i going wrong. Thanks
 A: The potential energy of the capacitor before the insertion of the dielectric slab is $$ U_0=\frac{C_0 V^2}{2} \tag{1}$$ with $$C_0=\frac {\epsilon_0 w l}{d}$$ The potential energy of the capacitor with the slab inserted a distance $x$ is approximately $$U(x)=V^2 \frac {w \epsilon_0 [(l-x)+\epsilon_r x]}{2d}=V^2 \frac {w \epsilon_0 [l+(\epsilon_r-1)x]}{2d}$$ When the slab is inserted a distance $x$, an additional charge $$Q_B(x)=V\frac {\epsilon_0 (\epsilon_r-1) w x}{d}$$ must be provided by the battery at potential $V$. Therefore, the (chemical) potential energy of the battery decreases by $Q_B(x)V$ $$U_B(x)=U_{B0}-Q_B(x)V=U_{B0}-V^2\frac {\epsilon_0 (\epsilon_r-1) w x}{d}$$ giving a total potential energy of the system battery plus capacitor $$U_{tot}(x)=U_B(x)+U(x)=U_{B0}+U_0 - V^2\frac {\epsilon_0 (\epsilon_r-1) w x}{2d}$$ Note that at an insertion length $x$ the potential energy of the capacitor has increased by $$\Delta U(x)=V^2\frac {\epsilon_0 (\epsilon_r-1) w x}{2d}$$ while the potential energy of the battery has decreased by twice as much $$\Delta U_B(x)=-V^2\frac {\epsilon_0 (\epsilon_r-1) w x}{d}$$ Therefore the total potential energy of battery plus capacitor becomes lower when the slab is inserted which means that a positive force is exerted on the slab $$F=-\frac {d U_{tot}(x)}{dx}=V^2\frac {\epsilon_0 (\epsilon_r-1) w}{2d}$$ It is important to note that this force performs mechanical work $W$ on the dielectric slab $$W=Fx=V^2\frac {\epsilon_0 (\epsilon_r-1) w x}{2d}$$ corresponding to the loss in potential energy of the total system so that the total energy of the system is conserved. 
This work can be converted into any other form of kinetic or potential energy permitted by energy conservation. It can be kinetic energy of the slab, it can be mechanical energy transferred to your hand during slow insertion, it can be potential energy of a spring or the lifting of a weight using a pulley, or it can be electromagnetic energy. It can also be transformed into heat, e.g. by friction or electrically. Thus, in general, heat is only one energy form into which the potential energy loss of the system battery + capacitor can be converted. In principle, the mechanical work performed by the system on the dielectric slab can be converted into any form of energy. It depends on the experimental setup whether the speed of the insertion could have an influence on the conversion of the performed work into heat.  
