A parallel-plate capacitor is made from two plates 13.0 c m on each side and 4.2

Question

A parallel-plate capacitor is made from two plates 13.0cm on each side and 4.20mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.50. (See the figure below(Figure 1) .) An 15.5V battery is connected across the plates.

A parallel-plate capacitor is made from two plates 13.0cm on each side and 4.20mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.50. (See the figure below(Figure 1) .) An 15.5V battery is connected across the plates. What is the capacitance of this combination? 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=L^2 =0.13^2

A=0.0169 m^2

C=C_air+C_plexigals

C =Keo(A/2)/d +eo(A/2)/d

C =(K+1)(eoA/2d)

C=(3.5+1)(8.85*10^-12*0.0169/2*4.2*10^-3)

C=8.01*10^-11 F

b)

E-=(1/2)CV^2 =(1/2)*(8.01*10^-11)*15.5^2

E=9.62*10^-9 J

c)

if we plexigals then capacitance

C=eoA/d =(8.85*10^-12)*0.0169/(4.2*10^-3)

C=3.56*10^-11 F

Energy stored

E=(1/2)CV^2=(1/2)*(3.56*10^-11)*15.5^2

E=4.28*10^-9 J

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