Two conducting plates, each with area 10 cm2, are separated a distance d = 6mm,

Question

Two conducting plates, each with area 10 cm2, are separated a distance d = 6mm, A dielectric slab with constant ic = 4 is inserted between them tilling two thirds of the space in between as in the figure. The plates have been charged as shown with Q = 4 mu C. Calculate the magnitude of the electric field E0 in the empty region between the plates. According to (a) calculate the potential Vab = Va- Vb between the left plate and the left face of the dielectric. Calculate the magnitude of the electric field E inside the dielectric. According to (c) calculate the potential Vbc= Vt - Vc between the left face of the dielectric and the right plate. Calculate the capacitance of this configuration.

Explanation / Answer

A) E0 = q/Aepsilon0 = 4e-6/(10e-4*8.85e-12) = 4.5198E8 = 4.52 * 10^8 N/C

B) Vab = Va - Vb = E d/3 = 4.5198e8 * (6e-3/3) = 9.0396e5 V = 9.04 * 10^5 V

C) E = q/Akepsilon0 = 4e-6/(10e-4*4*8.85e-12) = 1.13E8 = 1.13 * 10^8 N/C

D) Vbc = Vb - Vc = E 2d/3 = 1.13e8 * 2*6e-3/3 = 4.52e5 = 4.52 * 10^5 V

E) C = Q/V = Q/(Vab + Vbc) = 4e-6/(9.0396e5 + 4.52e5) = 2.95e-12 = 2.95 * 10^-12 F

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