Two parallel-plate capacitors, C_1 and C_2, are connected as shown at the right.

Question

Two parallel-plate capacitors, C_1 and C_2, are connected as shown at the right. Their dimensions are identical, but C_2 has a sheet of Teflon, with a dielectric constant of 2.1, that fills the region between its plates. The capacitance C_1 is labelled as 5.0 pF. (a) What is the equivalent capacitance between points a and b? Suppose I use a 90-V battery to charge the combination through a 100-M Ohm resistance, as shown at the right. How long after closing the switch S would I have to wait before the voltage difference between points a and b reaches 80 V?

Explanation / Answer

part a:

capacitance of a capacitor increases by a factor of dielectric constant when a dielectric is inserted in between its plates.

so capacitance of C2=2.1*capacitance of C1=2.1*5 pF=10.5 pF

as C1 and C2 are connected in series, total capacitance of the connection

=C1*C2/(C1+C2)

=(5*10.5)/(5+10.5)

=3.387 pF

part b:

time constant=resistance*equivalent capacitance

=100*10^6*3.387*10^(-12) s

=3.387*10^(-4) seconds

then voltage across the capacitor at any time t is given as

90*(1-exp(-t/time constant))

let at time t seconds, it reaches 80 volts

==>80=90*(1-exp(-t/(3.387*10^(-4))

==>8/9=1-exp(-t/(3.387*10^(-4))

==>1/9=exp(-t/(3.387*10^(-4))

==>t=-3.387*10^(-4)*ln(1/9)

=7.442*10^(-4) seconds

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