Question

A parallel plate capacitor consists of two circular metal plates, each of radius 1.38 cm. A coating of Teflon 40 m thick is applied to the inner surface of one plate to provide a dielectric layer, and then the plates are pressed together. Find the voltage to be applied between the plates of this capacitor to establish a charge of 0.5 nC on each plate. (Given: and dielectric constant of Teflon = 2).​

Answer 1

Answer:

may be this can help

Explanation:

r=1.38cm=1.38×10^-2 m

d=40×10^-6

Q= 0.5×10^-9

A=πr^2

=3.14×(1.38×1.38×10^-2)^2

=5.96×10^4

capacitance C = KۥA/d

=2×8.854×10^-12×5.96×10^-4/40×10^-6

=0.26×10^-9F

V=Q/C

=0.5×10^-9/0.26×10^-9

= 1.92 V

Answer 2

Answer:

The voltage to be applied between the plates of this capacitor to establish a charge of 0.5 ηC on each plate is 1.92 Volts.

Explanation:

Given-

$Radius,r=1.38cm=1.38\times10^{-2} m\\Diameter,d=40\times10^{-6}\mu m\\Charge,Q= 0.5\times10^{-9} C$

Now, calculate the area:

$A=\pi r^{2}$

$=3.14\times(1.38\times10^{-2})^2$

$=5.96\times10^4$ $m^{2}$

The area of the plate is $5.96\times10^4$ $m^{2}$.

Find the Capacitance:

$C = K$∈$\times A/d$

$=2\times8.854\times10^{-12}\times5.96\times10^{-4}/40\times10^{-6}\\=0.26\times10^{-9}F$

The capacitance is $0.26 \times 10^{-9} \mathrm{~F}$.

We will now finally calculate the Voltage with the help of Electric charge and Capacitance:

$V=Q/C\\=0.5\times10^{-9}/0.26\times10^{-9}\\= 1.92 V$

Hence, the voltage is 1.92 volts.