6. 110 points Tipler6 24.P.069 My Notes The positively charged plate of a parall

Question

6. 110 points Tipler6 24.P.069 My Notes 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 3.2 x 105 V/m. When the space is filled with a certain dielectric material, the field strength between the plates is reduced to 1.3 x 105 V/m (a) What is the dielectric constant of the material? (b) If Q = 11 nC, what is the area of the plates? cm2 (c) What is the total induced bound charge on either face of the dielectric material? nC eBook

Explanation / Answer

Eo = 3.2*10^5 V/m

E = 1.3810^5 V/m

dielectric constant k = Eo/E = (3.2*10^5)/(1.3*10^5) = 2.46

(b)

Eo = Q/A*eo

3.2*10^5 = (11*10^-9)/(A*8.84*10^-12)

A = 38.9 cm^2

(c)

q = 11 nc

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