A parallel-plate capacitor is made from two square plates 12 cm on each side sep

Question

A parallel-plate capacitor is made from two square plates 12 cm on each side separated a distance of 4.5 mm. I"hc space between the plates is initially empty. The capacitor is connected in series with a 10 MOhm resistor and an 18 v battery Find the capacitance of the capacitor AND the time constant of the circuit Next, half of the space between the plates, in the region shown by the dashed box, is filled with Plexiglass, which has a dielectric constant K = 3 4, The other half is left empty Calculate the new capacitance AND time constant of the circuit. (You can treat each half of the capacitor as an individual capacitor. Are they hooked together in series or parallel?) When the switch at S is closed, what will the voltage across the capacitor be after 2.5 time constants has passed? What percent of fully charged is this?

Explanation / Answer

capacitor=epsilon*Area/distance between plates

area=0.12*0.12=0.0144 m^2

distance between plates=4.5 mm=0.0045 m

capacitor=2.8*10^(-11) F

time constant=resistance*capacitance=2.8*10^(-4) sec

if dielectric is inserted

capaciotr on non dilectric space=2.8*10^(-11)/2=1.4*10^(-11) F

capacitance of dielectric space with K=3.4,=>3.4*1.4*10^(-11) F=4.76*10^(-11) F

new equivalent capacitance=1.4*10^(-11+4.76*10^(-11)=6.16*10^(-11) F

time constant=6.16*10^(-4) sec

(c) voltage across capacitor is given as

Vc=18(1-e^(-t/time contant)

at t=2.5*time constant

Vc=16.52 V

percentage=16.52*100/18=91.8 %

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