4. A parallel plate capacitor having an area A and plate separation d is charged

Question

4. A parallel plate capacitor having an area A and plate separation d is charged to a potential V by a battery. There is air between the plates. The battery is then disconnected and a dielectrid slab of thickness t and dielectric constant x is placed in between the plates as shown in the Figure below. Take A = 50 cm2, d = 1 cm, t = 0.5 cm, V-50 V, and K = 5.0. a) What is the capacitance prior to adding the dielectric slab? After the dielectric slab is introduced: b) How much free charge is on each plate? c) What is the electric field strength in the gap between the plates of the capacitor and the dielectric slab? d) What is the electric field strength in the dielectric? e) What is the potential difference between the plates? f) What is the capacitance of the parallel plate capacitor with the slab? Lastly, g) What percentage of the energy stored in the capacitor is stored in the air gaps and 10 What percentage is stored in the slab?

Explanation / Answer

A) With only air between the plates, The capacitance is given by

CO = Eo*A / D where Eo = permittivity of free space , A - area, D - separation between plates.

CO = 8.85e-12*50e-4 / 1e-2

CO = 4.4 pF or 4.4e-12 F

After the dielectric is inserted, the capacitance will increase by value of k which is given as 5. It will become 22 pF

B) Q = CO*VO = 4.4E-12*50 = 2.2E-10 Coulombs.. After the dielectric is inserted, the charge will stay the same as capacitor was disconnected.

C) E in the gaps = Q / EOA

E = 2.2E-10 / 8.85E-12*50E-4

E = 4971.75 N/C

D) E IN THE DIELECTRIC = 1/K * E in the gaps

= 1/5 * 4971.75

= 994.35 N/C

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