A parallel-plate air capacitor has rectangular plates, each of ar… - Vidyadip

Question

A parallel$-$plate air capacitor has rectangular plates, each of area $20 \ cm ^2$ separated by a distance of $2 \ mm$. The potential difference between the plates is $500$ volts. Calculate $(i)$ its capacitance $(ii)$ the charge on each plate $(iii)$ the electric field intensity between the two plates.

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Answer

Data : $A=20 \ cm ^2=20 \times 10^{-4} m ^2=2 \times 10^{-3} m ^2, k =1, V =500 V , d =2\ mm =2 \times 10^{-3}m , \varepsilon_0=8.85 \times 10^{-12} F / m$
$(i)$ Capacitance :
$C=\frac{A \varepsilon_0 k}{d}=\frac{\left(2 \times 10^{-3}\right)\left(8.85 \times 10^{-12}\right)(1)}{2 \times 10^{-3}}$
$=8.85 \times 10^{-12} F (=8.85 pF )$
$(ii)$ Charge:
$Q = CV =\left(8.85 \times 10^{-12}\right)(500)$
$=4.425 \times 10^{-9} C\ (=4.425\ nC )$
$(iii)$ Intensity of the electric field:
$E =\frac{V}{d}=\frac{500}{2 \times 10^{-3}}=2.5 \times 10^5 V / m$

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