Question

Consider the situation shown in figure (31-E29). The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf ε. All surfaces are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.
Figure

Answer 1

The value of M for which the dielectric slab will stay in equilibrium =

$M=\frac{\varepsilon_{0} b E^{2}(k-1)}{2 d g}$

Explanation:

In general we know that, the electric field attracts the dielectric into capacitor with minimal force

$\frac{\varepsilon_{0} \mathrm{b} \mathrm{V}^{2}(\mathrm{k}-1)}{2 \mathrm{d}}$

Where  b - width of the plates

             k – Dielectric constant  

            d – Separation between plates  

            V = E = Potential difference

Hence from the above fig, we can consider that surfaces are friction less, and this force is counteracted by weight .  

So,  $M=\frac{\varepsilon_{0} b E^{2}(k-1)}{2 d g}$