The positively charged plate of a parallel-plate capacitor has a charge equal to

Question

The positively charged plate of a parallel-plate capacitor has a charge equal to Q. When the space between the plates is evacuated of air, the electric field strength between the plates is 3.6 105 V/m. When the space is filled with a certain dielectric material, the field strength between the plates is reduced to 9.0 104 V/m. (a) What is the dielectric constant of the material? (b) If Q = 17 nC, what is the area of the plates? cm^2 (c) What is the total induced bound charge on either face of the dielectric material? nC

Explanation / Answer

A.The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space. It is an expression of the extent to which a material concentrates electric flux, and is the electrical equivalent of relative magnetic permeability.

B.

C = e0*area/separation
V = E*separation
Q = VC
We don't know the separation, but we can deduce area from the above.
Q = VC = E*separation*e0*area/separation = E*e0*area => area = Q/(E*e0)
I get area = 6.85705 E-3 m^2.
C. The free charge surface density (free) = e0*E(0), where E(0) is the vacuum field.
The bound charge density (bound) = -(free)*(1-(1/Kd)).
The bound charge = (bound)*area.
by using this formula you can calculate the induced bound charge on either face of the dielectric material.

the field strength between the plates is reduced to 9.0 10^4 V/m is wrong please provide right data.

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