Question

A parallel plate capacitor is given a charge of 5x10-8C. The space between the plates is filled with dielectric having a permittivity of 5.43x10-11F/m and the electric field within the dielectric is 5.81x106V/m. What is its plate area?

Answer 1

We find the area of the capacitor if we consider that the formula that relates the charge and area of the plates with the electric field generated:

$E=\frac{Q}{\epsilon_rA} \implies A=\frac{Q}{\epsilon_rE}=\frac{(5\times10^{-8}\,C)}{(5.43\times10^{-11}F/m)(5.81\times10^{6}V/m)}$

$A=\frac{Q}{\epsilon_rE}=1.585\times10^{-4}\,m^2$

In conclusion, the area of the plates is about 1.585X10^-4 m².

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.