Question

A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.70 cm wide and 12.0 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick.
What is the maximum charge that can be stored in this capacitor? (The dielectric constant of mica is 5.4, and its dielectric strength is 1.00 \times 10^8 \;{\rm{V}}/{{\rm{m}}})

Qmax= ??mC

Answer 1

Answer:

2.121876 mC

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $0.037\times 12\ m^2$

d = Thickness = 0.0225 mm

E = Dielectric strength = $1\times 10^8\ V/m$

k = Dielectric constant = 5.4

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}\\\Rightarrow C=\dfrac{5.4\times 8.85\times 10^{-12}\times 0.037\times 12}{0.0225\times 10^{-3}}\\\Rightarrow C=9.43056\times 10^{-7}\ F$

Maximum voltage is given by

$V_m=E_md\\\Rightarrow V_m=1\times 10^8\times 0.0225\times 10^{-3}\\\Rightarrow V_m=2250\ V$

Maximum charge is given by

$Q_m=CV_m\\\Rightarrow Q_m=9.43056\times 10^{-7}\times 2250\\\Rightarrow Q_m=0.002121876\ C=2.121876\ mC$

The maximum charge that can be stored in this capacitor is 2.121876 mC