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 65 % of the magnitude of the free charge on each plate. The capacitance is 4900, where is a constant with dimensions of length, and the maximum charge that can be stored on the capacitor is 21020Emax, where Emax is the breakdown threshold.

Part A Part complete

What is the dielectric constant for the slab?

= 2.9

What is the plate separation distance in terms of ?

d =0.43  

Part C

What is the plate area in terms of ?

A = 2  

Explanation / Answer

Solution:-

Qind = (1 - 1/k)*Q
0.65 = (1 - 1/k)

=>-0.35= -1/k

=>k=1/0.35

k = 2.86

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

also, Q = CV = 450e*V/l = 490e*Emax*d/l = 210*l^2*e*Emax
so, 490d= 210l^3
d = 0.43l^3

Ak/d = 450/l

Ak=(450/l)*d

=>A=((450/l)*d)/k

=> ((450/l)* 0.43l^3)/2.86

=>A= 193.5l^2/2.86
A = 67.66l^2

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