An electric field given by E^rightarrow = 1.5 i^cap = 8.8 (y^2 + 9.1) j^cap pier

Question

An electric field given by E^rightarrow = 1.5 i^cap = 8.8 (y^2 + 9.1) j^cap pierces the Gaussian cube of edge length 0.170 m and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? (a) Number ________ Units (b) Number ________ Units (c) Number ________ Units (d) Number ________ Units (e) Number ________ Units

Explanation / Answer

Solution:

A)

Top face:

flux = E(y)*A = 8.8(0.17^2+9.1) * (0.17)^2

=> flux = 2.32166 Wb

B)

Bottom Face:

flux = E(y)*A = 8.8(0^2+9.1) * (0.17)^2

=> flux = 2.3143 Wb

C)

Left Face :

flux = E(x)*A = 1.5 * 0.17^2

=> flux = 0.04335 Wb

D)

Back Face:

flux = E(z)*A = 0 * 0.17^2 = 0 Wb

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