Two parallel plates are .005m apart and are each 2nr in area. The plate are in v

Question

Two parallel plates are .005m apart and are each 2nr in area. The plate are in vacuum and an electric potential difference of 10,000V is applied across them. Find the capacitance, the charge on each plate, the electric field intensity in the space between the plates, and the stored energy. If a dielectric material with dielectric constant k=80.4 is inserted into the gap between the plates, with the electric potential remaining the same, find the new capacitance, charge on each plate, electric field intensity, and energy stored.

Explanation / Answer

A) for plate capacitor

C=0*A/d=0*2/0.005=3.54e-9 F=3.54 nF

Q=V*C=3.54e-5 C

E=V/d=10000/0.005=2e6 V/m

B)K=80.4

C=k*0*A/d=0*80.1*2/0.005=2.87e-7 F

Q=V*C=2.87e-3 C

E=V/d=2e6 V/m

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