Question

An electric field of 8.50 times 10^5 V/m is desired between two parallel plates, each with an area of 35.0 cm^2 separated by 3.00 mm of air. What charge must be on each plate? How does the energy stored on a parallel plate capacitor change if: The potential difference applied between the plates is doubled? The charge on each plate is doubled? The separation between the plates is doubled, as the capacitor remains connected to the same battery? The separation between the plates is halved, and the difference of potential between the plates is doubled_____________During a typical lightning bolt, about 20 C of charge are transferred from the charged clouds to the ground. It has been proposed to collect and store this "free electricity" for further use in order to reduce our dependence on fossil fuels. Assume you were to collect this charge on a parallel plate capacitor filled with paper with a dielectric constant of 3.7 and dielectric strength of 16 times10^6 V/m. What would be the minimum area of the plates of this capacitor? If the thickness of the capacitor is 5 cm (plates, dielectric and supporting frame), what is the volume of this storage unit? Given that the physics lab is about 10m times 15m times 2m, how does the volume of the capacitor compare to that of the lab?__________

Answer 1

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The electric field and potential difference between the plates are related as, Here d is the distance betw een the plates, If qis the charge on each plate then The capacitance of a capacitor with plate area Ά1s C-pt. Thus, This implies, - (8.85x1012) (35x104(8.50x105 = 2.63× 10-8 C