A parallel-plate capacitor, having plates of 100c㎡ and has three
dielectrics 1mm each and of permittivity’s 3, 4 and 6. If the peak
voltage 2000V is applied to the plates, calculate the energy stored
in each dielectric
V=100⋅10−4⋅1⋅10−3=10−5 (m3)V=100\cdot10^{-4}\cdot1\cdot10^{-3}=10^{-5}\ (m^3)V=100⋅10−4⋅1⋅10−3=10−5 (m3)
E=2000/0.003=666667 (V/m)E=2000/0.003=666667\ (V/m)E=2000/0.003=666667 (V/m)
E1=ϵ1ϵ0E2V/2=3⋅8.85⋅10−12⋅6666672⋅10−5/2=E_1=\epsilon_1\epsilon_0E^2V/2=3\cdot8.85\cdot10^{-12}\cdot666667^2\cdot 10^{-5}/2=E1=ϵ1ϵ0E2V/2=3⋅8.85⋅10−12⋅6666672⋅10−5/2=
=5.9⋅10−5 (J)=5.9\cdot10^{-5}\ (J)=5.9⋅10−5 (J)
E2=ϵ2ϵ0E2V/2=4⋅8.85⋅10−12⋅6666672⋅10−5/2=E_2=\epsilon_2\epsilon_0E^2V/2=4\cdot8.85\cdot10^{-12}\cdot666667^2\cdot 10^{-5}/2=E2=ϵ2ϵ0E2V/2=4⋅8.85⋅10−12⋅6666672⋅10−5/2=
=7.84⋅10−5 (J)=7.84\cdot10^{-5}\ (J)=7.84⋅10−5 (J)
E3=ϵ3ϵ0E2V/2=6⋅8.85⋅10−12⋅6666672⋅10−5/2=E_3=\epsilon_3\epsilon_0E^2V/2=6\cdot8.85\cdot10^{-12}\cdot666667^2\cdot 10^{-5}/2=E3=ϵ3ϵ0E2V/2=6⋅8.85⋅10−12⋅6666672⋅10−5/2=
=11.8⋅10−5 (J)=11.8\cdot10^{-5}\ (J)=11.8⋅10−5 (J)
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments