Shown a parallel plate capacitor with plates of width `b` and length `l`. The separation between the platesis d. The plates are rigidly clamped and connected to a battery of emf V. A dielectric slab of thickness d and dielectric constant K is slowly inserted between the plates. (a) Calculate the energy of the system when a length x of the slab is introduced into the capacitor. (b) what force should be applied on teh slab to ensure that is goes slowly into the capacitor? Neglect any effect of friction or gravity.

Explanation

The plate area of the part with the dielectric is bx. Its capacitance is `
` C_1 = (K epsilon_0 bx)/(d)`
`. Similarly, the capacitance of the part without the dielectric is `
`C_2 = (epsilon_0 b)/(d) [ l + x (K- 1)]. The energy of the capacitor is `
`U = (1)/(2) CV^2`
` `
`= (epsilon_0 bV^2)/( 2 d) [ l + x(K-1)]`
`. (b) Suppose, the electric field attracts the dielectric slab with a force F. An external force of equal magnitude F should be applied in opposite direction so that the plate moves slowly ( no acceleration). Consider the part of motion in which the dielectric moves a distance dx further inside the capacitor. The capacitance increases to C+dC. As the potential difference remains constant at V, the battery has to supply a further charge `
` dQ = (dC)V`
` to the capacitor. The work done by the battery is, therefore, `
`dW_b = V dQ = (dC)V^2`
`. The external force F does a work `
`dW_e = (-F dx)`
` during the displacement. The total work done on the capacitor is `
` dW_b +dW_e = (dC)V^2 - Fdx`
`. this should be equal to the increase dU in the stored energy. thus, (1)/(2) (dV)V^2 = (dC)V^2- Fdx`
`. or, F = (1)/(2) V^2 (dC)/(dx). Using equation (i) `F= (epsilon_(0) b V^(2) (K-1))/((2d)`
Thus, the electric field attracts the dielectric into the capacitor with a force `(epsilon_(0) b V^(2) (K-1))/(2d)` and this much force should be applied in the opposite direction.