A cubic box of Side a = 0.320 m is placed so that its edges are parallel to the

Question

A cubic box of Side a = 0.320 m is placed so that its edges are parallel to the coordinate axes, as shown in the figure. There is NO net electric charge inside the box, but the space in and around the box is filled with a no uniform electric field of the following form: E(x, y, z) = Kz j + Ky k, where K = 3.30 N/(Cm) is a Constant. What is the electric flux through the top face of the box? (The top face of the box is the face where z = a. Remember that we define positive flux pointing out of the box.) What is the total electric flux through the five other faces of the box? (Again, outward flux is positive.)

Explanation / Answer

step;1

Given that

the electric field E=Kxi+Ky j

step;2

now we find the the electri flux throught top face the box

here Z=a

the electric field E=3.3*a i+0 j=3.3 a

the electric flux E=3.3a^3=3.3*0.32^3=0.11 Nm^2/C

step;3

now we find the electric flux on remaining five sides

the electric field Enet =Ebottom+Eleft side+Eright side +E top right +E left right

                                =0+(Ka j+ka k)+(Ka j+Ka k)+Ka k+Ka j

                               =(3Ka j+3ka k )

                                =3Ka

                                =3*3.3*0.32

                                  =3.2 N/c

the total electric flux through five other surface =3.2*0.32^2=0.33 N.m^2/C

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