Answer to Question #212527 in Electric Circuits for Theo

Question #212527

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?


1
Expert's answer
2021-07-05T08:44:48-0400

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=QϵrA    A=QϵrE=(5×108C)(5.43×1011F/m)(5.81×106V/m)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=QϵrE=1.585×104m2A=\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.

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