Dielectric Removal from Charged Capacitor – Rankers Physics
Topic: Capacitors
Subtopic: Capacitor With Dielectrics

Dielectric Removal from Charged Capacitor

The capacitance of a parallel plate capacitor is (C) when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant (k). The capacitor is connected to a cell of emf (E), and the slab is taken out
charge (CE(k-1)) flows through the cell
energy (E^2 C(k-1)) is absorbed by the cell.
the energy stored in the capacitor is reduced by (E^2 C(k-1))
the external agent has to do (frac{1}{2} E^2 C(k-1)) amount of work to take the slab out.

Solution:

Initially, with dielectric (k) and connected to emf (E), capacitance is (C_i = kC) and energy is (U_i = frac{1}{2} kCE^2). When the slab is taken out while connected to the cell, capacitance becomes (C_f = C) and energy is (U_f = frac{1}{2} CE^2). The change in energy is (Delta U = U_f - U_i = frac{1}{2} CE^2 (1-k) = -frac{1}{2} (k-1)CE^2). The charge flowing into the cell is (Delta Q = Q_i - Q_f = (kC - C)E = (k-1)CE). Work done by the cell on the capacitor is (W_{cell} = -E Delta Q = -E^2 C(k-1)). By work-energy theorem, (W_{ext} + W_{cell} = Delta U). Thus, (W_{ext} = Delta U - W_{cell} = -frac{1}{2} (k-1)CE^2 - (-E^2 C(k-1)) = frac{1}{2} (k-1)CE^2).

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