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# Energy stored in capacitor

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Consider a parallel plate capacitor with area A and separation d. The plates have fixed charges Q and -Q, and no battery connected. Determine the energy stored before and after the insertion of a dielectric with dielectric constant K that completely fills the space between the plates.

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SOLUTION

Fundamentals in a nutshell

1. Capacitance of a parallel plate capacitor with air between the plates C = Є0A/d where Є0 = permittivity of air, A = area of each plate, d= separation between the plates.

2. Dielectric constant K of a dielectric medium with permittivity Є is defined as : K = Є/Є0.

3. Capacitance of a capacitor with a dielectric slab (dielectric constant K) between the plates (filling the space completely) is given by C' = ЄA/d = KЄ0A/d = KC where C = capacitance with air between the plates.

4. Energy stored in a capacitor of capacitance C with charge +Q on one plate and -Q on the other and potential difference V between the plates is given by E = Q2/2C = ½ CV2

In the given problem the capacitor is charged to +Q and -Q with air as medium between the plates. Thus, the energy stored is given by : E = Q2/2C = Q2d/2Є0A ...........(1)

When a dielectric slab with permittivity Є (dielectric constant K) is inserted between the plates completely filling the space, its capacitance becomes C' = KЄ0A/d. Therefore, the energy stored with dielectric slab = E' = Q2d/2ЄA = Q2d/2KЄ0A = E/K ..........(2)

It should be noted that the charge Q remains unchanged as the capacitor is isolated (not connected to battery) and the charge is trapped on the plates.

From (1) and (2) it is clear that with the dielectric slab between the plates, the energy stored is reduced by a factor of 1/K as compared to the energy stored with air in between. Where does the difference in energy in the two cases go? As the slab is brought near the capacitor for insertion between the plates, a pull is experienced on the slab pulling it in between the plates spontaneously. This means the capacitor is doing work on the slab in pulling it between the plates. The energy difference goes to do this work on the slab.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!