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.

(b) If Q = 19 nC, what is the area of the plates? in (cm^2)

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

Thank you very much in advance

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 =19*10^-9/(3.5*10^5*8.85*10^-12) =6.13810^-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. I get Kd = E(0)/E(d) = 3.5/1.5 = 2.33, ?(free) = 8.85*10^-12*3.5*10^5 =3.1*10^-6 C/m^2, ?(bound) =..... C/m^2, Q(bound) = ... C. use above results n get the answers

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