A capacitor consists of two stationary plates shaped as a semi-circle of radius `R` adn a movable plate made of dielectric with permittivity `epsilon` and capacble of rotoning about an axis `O` between the stationary plates (fig). The thickness of the movables plate is equal to `d` which is practically the separation between teh stationary plates. A potential difference `V` is applied to the capacitor. FInd the magnitude of the moment of forces relative to the axis `O` acting on teh movable plate in the position shown in the figure.

Explanation

When the capacitor is kept at a contant potential difference `V`, the work performed by the moment of electrostatic forces between the plates when the inner moveable plate is rotated by an angel `d varphi` equals the increases in the potential energy of the system. This comes about because when chagres are made, chagres flow from the battery to keep the potential constant and the amount of the work done by these charges is twice it magnitude and opposite in sign to the change in the energy of the capacitor Thus
`N_(z) = (del U)/(del varphi) = (1)/(2) V^(2) (del C)/(del varphi)`
Now the capacitor can be thought of as made up two parts (with and without the dielectric) in parallel.
Thus `C = (epsilon R^(2) varphi)/(2d) + (epsilon_(0) epsilon (pi - varphi) R^(2))/(2d)`
as the area of a sector of angle `varphi` is `(1)/(2) R^(2) varphi`. Defferenctian then gives
`N_(s) = - ((epsilon - 1) epsilon_(0) R^(2) V^(2))/(4d)`
The negative sign of `N_(x)` indicates that the moment of the force is acting clockwire (i.e., trying to suck in the-dielectric).