Practice Problemm Physics 272 Recitation, Spring 2018 Consider an ideal parallel

Question

Practice Problemm Physics 272 Recitation, Spring 2018 Consider an ideal parallel plate capacitor, having vacuum between two plates of area A and separation d. The shape of the plates is arbitrary. By the following steps, prove that the capacitance is where Eo is the permittivity. Assume that the charge on each plate is ±Q, and that the electric field E between the plates is uniform. 1. Use Gauss' law to calculate the electric field E between the plates in terms of Q and A. (Use a "Gaussian pillbox" that just encloses one of the plates, and remember that the field on the outside is zero.) 2. Use your expression for E to calculate the voltage V across the plates in terms of Q A, and d. 3. Calculate the capacitance C from the definition Q-CV.

Explanation / Answer

from the gauss's law, we know that

E*A = Q/e0

and electric field and voltage across plates

E =V/d

now put value of E in above equation

V*A/d = Q/e0.................. eq1

and we have that Q = C*V

so put value of Q in eq1

V*A/d = C*V/e0

V will be cancle out

A/d = C/e0

C = A*e0/d.....[proved]

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