A dielectric slab completely fills the space between the plates of a parallel-pl

Question

A dielectric slab completely fills the space between the plates of a parallel-plate capacitor. The magnitude of the bound charge on each side of the slab is 70 % of the magnitude of the free charge on each plate. The capacitance is 450(0), where is a constant with dimensions of length, and the maximum charge that can be stored on the capacitor is 2502(0)Emax, where Emax is the breakdown threshold.

Part A) What is the dielectric constant for the slab?

Part B) What is the plate separation distance in terms of ?

Part C) What is the plate area in terms of ?

Explanation / Answer

Qind = (1 - 1/k)*Q
0.7 = (1 - 1/k)
k = 3.333

C = Ake/d = 450e/l
so, Ak/d = 450/l

also, Q = CV = 450e*V/l = 450e*Emax*d/l = 250*l^2*e*Emax
so, 450d= 250l^3
d = 0.555l^3

A = 75l^2

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