A parallel-plate capacitor is made from two plates 12.5cm on each side and 4.45m
ID: 2287719 • Letter: A
Question
A parallel-plate capacitor is made from two plates 12.5cm on each side and 4.45mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.15. (See the figure below(Figure 1) .) An 20.0V battery is connected across the plates.
A.)What is the capacitance of this combination?
B.)How much energy is stored in the capacitor?
C.)If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?
Explanation / Answer
a)
Area
A=S2=0.1252=0.015625 m2
Capacitance
C=eo(A/2)/d+Keo(A/2)/d
C=(eoA/2d)(K+1) =[(8.85*10-12)(0.015625)/(2*(4.45*10-3)](3.15+1)
C=6.45*10-11 F or 64.5 pF
b)
Energy stored in capacitor is
E=(1/2)CV2=(1/2)*(6.45*10-11)*202
E=1.29*10-8 J or 12.9 nJ
c)
Capacitance without dielectric
C=eoA/d=(8.85*10-12)*0.015625/(4.45*10-3)
C=3.11*10-11F
Energy stored
E=(1/2)(3.11*10-11)*202
E=6.21*10-9 J or 6.21 nJ