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A cube has sides of lengthL=0.300mm. One corner is at the origin (Fig. E22.6). The nonuniform electric field is given byE=(-5.00N/C×m)x+(3.00N/C×m)zk.(a) Find the electric flux through each of the six cube facesS1,S2,S3,S4,S5andS6. (b) Find the total electric charge inside the cube.

Short Answer

Expert verified

(a)

The electric flux through the first side of the cube face surface is0Nm2/C.
The electric flux through the second side of the cube face surface is0.081Nm2/C
The electric flux through the third side of the cube face surface is0Nm2/C
The electric flux through the fourth side of the cube face surface is0Nm2/C
The electric flux through the fifth side of the cube face surface is-0.135Nm2/C
The electric flux through the sixth side of the cube face surface is0Nm2/C

(b)

The total electric charge inside the cube is-4.781×10-13C.

Step by step solution

01

Define free body diagram and given data

The concept of electric flux, its calculation, and the analogy between the flux of an electric field and that of water. Let us imagine the flow of water with a velocityvin a pipe in a fixed direction, say to the right. If we take the cross-sectional plane of the pipe and consider a small unit area given by ds from that plane, the volumetric flow of the liquid crossing that plane normal to the flow can be given as vds.

02

Simplify the electrical flux

As equation22.8mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by:

ϕE=Qenclosedε0

As equation 22.5mentions, the electric flux through a surface is given by

localid="1665115999561" ϕE=EAcosf=E×dA

Solve for the first side of the cube faceS1:

ϕS1=E×nS1A=E×^A=0

So, the electric flux through the first side of the cube face surface is0Nm2/C.

Solve for the second side of the cube faceS2:

localid="1665116435683" ϕS2=E×n^S2A=E×k^A=3N/C×m×z×0.3m2=3N/C×m×0.3m×0.3m2=3N/C×m×0.3m3=0.081N×m2/C

So, the electric flux through the second side of the cube face surface is0.081Nm2/C.
Solve for the third side of the cube faceS3:

ϕS3=E×n^3A=E×^A=0N×m2/C

So, the electric flux through the third side of the cube face surface is0Nm2/C

Solve for the fourth side of the cube faceS4:

ϕS4=E×n^S4A=E×k^A=-3N/C×m×z×0.3m2=-3N/C×m×0m×0.3m2=0N×m2/C

So, the electric flux through the fourth side of the cube face surface is0Nm2/C.
Solve for the fifth side of the cube faceS5:

ϕS5=E×n^S5A=E×l^A=-5N/C×m×x×0.3m2=-5N/C×m×0.3m×0.3m2=-5N/C×m×0.3m3=-0.135N×m2/C

So, the electric flux through the fifth side of the cube face surface is -0.135Nm2/C.

Solve for the sixth side of the cube faceS6:

ϕS6=E×S^6A=-E×ι^A=-5N/C×m×x×0.3m2=-5N/C×m×0m×0.3m2=0N×m2/C

So, the electric flux through the sixth side of the cube face surface is0Nm2/C.

03

Solve total electric charge

Solve for the total electric charge inside the cube:
We will take the data (electric flux through each side of the cube) from part (a). In order to evaluate the total electric charge density inside the cube, we use the following relation:

Φtotal=ΦS=ΦS1+ΦS2+ΦS3+ΦS4+ΦS5+ΦS6=0+0.081+0+0+-0.135+0Nm2/C=-0.054Nm2/C

As equation 22.8mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by:

Φtotal==ΦS=ΦS1+ΦS2+ΦS3+ΦS4+ΦS5+ΦS6=0+0.081+0+0+-0.135+0Nm2/C=-0.054Nm2/C

So, the total electric charge inside the cube is-4.781×10-13C.

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