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 3.5 105 V/m. When the space is filled with a certain dielectric material, the field strength between the plates is reduced to 1.5 105 V/m.


(a) What is the dielectric constant of the material?


(b) If Q = 19 nC, what is the area of the plates?
. cm2

(c) What is the total induced bound charge on either face of the dielectric material?

Explanation / Answer

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 = 6.45377E-3 m^2. C. The free charge surface density s(free) = e0*E(0), where E(0) is the vacuum field. The bound charge density s(bound) = -s(free)*(1-1/Kd). The bound charge = s(bound)*area. I get Kd = 2.5454545, s(free) = 2.47917E-6 C/m^2, s(bound) = -1.50521E-6 C/m^2, Q(bound) = 9.71429E-9 C.

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