Answer:
289282
Explanation:
r = Radius of plate = 0.52 mm
d = Plate separation = 0.013 mm
A = Area = [tex]\pi r^2[/tex]
V = Potential applied = 2 mV
k = Dielectric constant = 40
[tex]\epsilon_0[/tex] = Electric constant = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]
Capacitance is given by
[tex]C=\dfrac{k\epsilon_0A}{d}[/tex]
Charge is given by
[tex]Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}[/tex]
Number of electron is given by
[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}[/tex]
The number of charge carriers that will accumulate on this capacitor is approximately 289282.
