Home
Class 12
PHYSICS
Show that the force on each plate of a p...

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(1//2)` QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor `1//2`.

Text Solution

Verified by Experts

Hint: Suppose we increase the separation of the plates by `Deltax` . Work done (by external agency) = F =`Deltax` . This goes to increase the potential energy of the capacitor by u a`Deltax` where u is energy density. Therefore, F = u a which is easily seen to be (1/2) QE, using `u = (1//2) epsilon_(0)E^(2)`. The physical origin of the factor 1/2 in the force formula lies in the fact that just outside the conductor, field is E, and inside it is zero. So, the average value E/2 contributes to the force.
Promotional Banner

Similar Questions

Explore conceptually related problems

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness (3/4)d, where is the separa - tion of the plates. How is the capacitance when the slab is inserted between the plates ?

A parallel plate capacitor has a capacitance of 10 mu F without dielectric. A dielectric of dielectric constant 2 is used to fill exactly half the thickness between the plates. The capacitance in mu F, now is

A closed loop moves normal to the constant electric field between the plates a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor plates (ii) When it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop.