A parallel-plate capacitor is made from two aluminum foil sheets, each 20 cm lon

Question

A parallel-plate capacitor is made from two aluminum foil sheets, each 20 cm long and 5.0cm wide. The two sheets are seperated by a distance of 0.015 mm.

a) what is the capacitance of this capacitor?

b) if a potential difference of 500 mV is applied across the two plates of this capacitor, what is the absolute value of the charges on each plate of the capacitor?

c) Now a piece of teflon with a dielectric constant of k= 2.1 is inserted to fill the space between the plate. What is the capacitance of the capacitor with the teflon inserted?

d) if a potential difference of 500 mV is applied as before across the two plates of this capacitor, what is the absolute value of the charges on each plate with the teflon inserted?

Explanation / Answer

Area

A=0.2*0.05 =0.01 m^2

Capacitance

C=eoA/d =(8.85*10^-12)*0.01/(0.015*10^-3)

C=5.9*10^-9 F or 5..9 nF

b)

Q=CV =(5.9*10^-9)*(500*10^-3)

Q=2.95*10^-9 C or 2.95 nC

c)

Ct =KC =2.1*5.9*10^-9

Ct=1.239*10^-8 F or 12.39 nF

d)

Q=12.39*500*10^-3

Q=6.195*10^-9 C or 6.195 nC

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