Please break down your steps and explaination please. 27.1 The electric flux ins

Question

Please break down your steps and explaination please.

27.1 The electric flux inside a parallel-plate capacitor Two 100 cm^2 parallel electrodes are spaced 2.0 cm apart. One is charged to +5.0 nC, the other to -5.0 nC. A 1.0 cm X 1.0 cm surface between the electrodes is tilted to where its normal makes a 45 degree angle with the electric field. What is the electric flux through this surface? MODEL Assume the surface is located near the center of the capacitor where the electric field is uniform. The electric flux doesn't depend on the shape of the surface.

Explanation / Answer

Here ,

surface charge density on plates , sigma = 5*10^-9 /100 *10^-4

sigma = 5 *10^-7 C/m^2

electric Field between the plates = sigma/epsilon

E = 5 *10^-7/8.854 *10^-12

E = 5.65 *10^4 N/C

Now , flux = E*A*cos(theta)

FLux = 5.65 *10^4 * 1 *10^-4 * cos(45)

Flux = 3.993 N.m^2/C

the electric flux through the surface is 3.993 N.m^2/C

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