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 80 % of the magnitude of the free charge on each plate. The capacitance is 4300, where is a constant with dimensions of length, and the maximum charge that can be stored on the capacitor is 29020 Emax, where Emax is the breakdown threshold.

Questions:

1. What is the dielectric constant for the slab?

2.What is the plate separation distance in terms of ?

3.What is the plate area in terms of ?

Explanation / Answer

Qind = (1 - 1/k)*Q
0.8= (1 - 1/k)
k = 5

C = Ake/d = 430e/l
so, Ak/d = 430/l

also, Q = CV = 430e*V/l = 430e*Emax*d/l = 290*l^2*e*Emax
so, 430d= 290l^3
d = 0.674*l^3

A = 79*l^2(approximately)

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