Question

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an insulating material with dielectric constant of 40. If a potential is applied to this device of 2.0 mV, how much charge will accumulate on this capacitor, in terms of the number of charge carriers?

Answer 1

289282

Step-by-step explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area =
`\pi r^2`

V = Potential applied = 2 mV

k = Dielectric constant = 40

`\epsilon_0` = Electric constant =
`8.854* 10^(-12)\ \text{F/m}`

Capacitance is given by

`C=(k\epsilon_0A)/(d)`

Charge is given by

`Q=CV\\\Rightarrow Q=(k\epsilon_0AV)/(d)\\\Rightarrow Q=(40* 8.854* 10^(-12)*\pi * (0.52* 10^(-3))^2* 2* 10^(-3))/(0.013* 10^(-3))\\\Rightarrow Q=4.6285* 10^(-14)\ \text{C}`

Number of electron is given by

`n=(Q)/(e)\\\Rightarrow n=(4.6285* 10^(-14))/(1.6*10^(-19))\\\Rightarrow n=289281.25\ \text{electrons}`

The number of charge carriers that will accumulate on this capacitor is approximately 289282.