A cubic cardboard box of side a = 0.290 m is placed so that its edges are parall

Question

A cubic cardboard box of side a = 0.290 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 nonuniform electric field of the following form: E(x,y,z) = Kz j + Ky k, where K= 3.60 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

side length of the box, a=0.29m

E(x,y,z)=(Kz)j+(ky)k and K=3.6 N/C

a)

electric flux =int(int(E.dA))

electric flux=int(int(E*cos(theta)dA))

electric flux=int(int(E*cos(theta)*dx*dy))

here,

E=sqrt((Kz)^2+(Ky)^2) and cos(theta)=Ky/(sqrt((Kz)^2+(Ky)^2))

now,

electric flux=int(int(E*cos(theta)*dx*dy))

electric flux=int(int(sqrt((Kz)^2+(Ky)^2)*Ky/(sqrt((Kz)^2+(Ky)^2))*dx*dy))

electric flux=int(int(Ky*dx*dy))

electric flux=k*int(int(y*dx*dy))

electric flux=k*int(int(dx)*ydy)

electric flux=k*int(x*ydy)

electric flux=k*x*(int(ydy)) ( limits are, x=0 to a and y=0 to a)

electric flux=k*x*(y^2/2)

electric flux=k*(a)*(a^2/2)

electric flux=1/2*k*a^3 ------is the electric flux throgh the top of the box ( one face)

and

electric flux=1/2*3.6*0.29^3

electric flux=0.044 N.m^2/C

b)

given, total electic flux throgh he entire box is zero,

hence,

through five other faces,

electric flux=electric flux=-1/2*k*a^3

or

electric flux=-0.044 N.m^2/C

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