Initially all the capacitor is uncharged and the switch S is open. The capacitor
ID: 1359566 • Letter: I
Question
Initially all the capacitor is uncharged and the switch S is open. The capacitor is a simple parallel plate capacitor made of two parallel conducting sheets with an air gap between them. IN what follows, assume the dielectric constant of air is exactly 1. With the switch closed, a dielectric slab of dielectric constant kappa > 1 is now inserted between the plates of the capacitor. With the dielectric in place, the switch S is now opened. The dielectric is now removed. What now is the stored charge? Has it changed? What now is the stored energy? Has it changed? Account for any change in the energy (in other words, if the energy changed, something did work: who or what?)Explanation / Answer
Let the initial Capacitance = C
Now Dielectric is inserted, Newe Capacitance = C*k
With Switch closed, With battery of EMF V volt.
Charge stored in Capacitor. Q = C*V
Q = C*k*V
Now, Dielectric is removed. But Charge will remaqin same = C*k*V
New Capacitance will be again = C
As the Charge Remains same,
V = Q/C
V = C*k*V/C
V = k*V
So the potential Difference increases.
Energy = 0.5*C*V^2
Energy = 0.5*C*(kV)^2 = k^2*C*V^2
The stored energy increases by the Factor of k^2
This happened because, External Agent did the work on removing the dielectric Slab from the Capacitor.