# Force required to insert dielectric in a capacitor connected to a battery [duplicate]

This is a sum in a reference book( H.C. Verma, concepts of phycics). A dielectric slab is being inserted in a parallel plate capacitor as shown in the diagram. The width of the plates is given to be $$b$$. We need to find the force exerted by the capacitor on the slab. The solution given in the book is as follows:

As voltage across the capacitor is same, the two sections are in parallel

$$\therefore c = c_1 + c_2 =\frac {\epsilon_0 b} {d} [l + x(k-1)]$$

$$U=\frac 1 2 cV^2$$

$$F=-\frac {dU}{dx} = -\frac {V^2}{2} \frac {dc}{dx} = -\frac {\epsilon_0 b V^2(K-1)}{2d}$$

I have some doubts about this solution. Firstly, the force is independant of $$x$$ which is a bit weird. We can see that the force is zero when $$x=l$$. Thus, the potential energy should have a maxima at $$x=l$$, which isn't seen here. Secondly, the charge on the plates is continuously changing, so current is flowing from the battery. Thus, there should be some heat that is generated, which is not taken into account.

• The formula for capacitance doesn't seem right, since it isn't symmetrical about $x = l$ which is automatically present in the question by symmetry. Commented Jul 28, 2020 at 19:15

Seems ok to me. The system has gained $$\frac{CV^2(K-1)}{2}$$ of energy and so that much work=$$Fl$$ must be done. $$C=\frac {\epsilon_0 b l}{d}$$. If the dielectric slab is moved into place slowly, there should be very little current, and given an ideal battery and negligible resistance otherwise, negligible heating. Regarding the x dependence, if the slab has moved a distance x the energy will increase by $$\frac{\epsilon_0(K-1)xV^2}{2d}$$ which then equals Fx. So F is still independent of the distance the slab moves.