A parallel-plate capacitor is made from two plates 12.0 c m on each side and 4.5

Question

A parallel-plate capacitor is made from two plates 12.0cm on each side and 4.50mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. (See the figure below(Figure 1) .) An 18.0V battery is connected across the plates.

1) How much energy is stored in the capacitor?

2)If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?

A parallel-plate capacitor is made from two plates 12.0cm on each side and 4.50mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. (See the figure below(Figure 1) .) An 18.0V battery is connected across the plates. How much energy is stored in the capacitor? If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?

Explanation / Answer

a)

Area A=0.12^2 =0.0144 m^2

Capacitance of the medium filled with air

C_air=eo(A/2)/2

C_air=(8.85*10^-12)(0.0144/2)/(4.5*10^-3)

C_air=1.416*10^-11 F

Capacitance of medium filled with plexigals

C_plex=Keo(A/2)/d

C_plex =KC_air =3.4*1.416*10^-11

C_plex=4.814*10^-11 F

total capacitance

C =C_air+C_plex

C=6.23*10^-11 F

total energy stored in capacitor is

E=(1/2)CV^2 =(1/2)*(6.23*10^-11)*18^2

E=1*10^-8 J or 10 nJ

b)

total capacitance is

C=eoA/d =(8.85*10^-12)*0.0144/(4.5*10^-3)

C=2.83*10^-11 F

E=(1/2)*(2.83*10^-11)*18^2

E=4.59*10^-9 J or 4.59 nJ

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