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dgoudie
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[SOLVED] Tough Flux question
A cubic cardboard box of side a = 0.250 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= 3.20 N/C.m is a constant.
http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype53/prob07a_cube.gif
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.)
Hint:Since E is not constant, you will need to do an area integral for this problem. Remember the definition of electric flux. Only one component of E contributes to the flux out of the top of the box. In doing the integral to find the flux, it might help to imagine breaking the top surface into strips along which the z-component of the electric field is constant. If all else fails, ask for help.
Flux=[tex]\int[/tex]E*dA
The Nonuniform part is what's tripping me up. I know how to do it with uniform electric field.
I'll call the top face faceA, Flux=I. So I[tex]_{A}[/tex]=[tex]\int[/tex]E*dA= [tex]\int[/tex](3.20 N/C z [tex]\widehat{j}[/tex] + 3.20 N/C y[tex]\widehat{k}[/tex])* (dA [tex]\widehat{k}[/tex])
So it says that for this face Z=the side length, but what do I do with y? and with the top face wouldn't only the K-hat vectors apply anyway?
and once I get an asnwer will it be negative?
anyway if someone could give me some insight, let me know if I've set it up right it would help a lot. Thanks in advance!
Homework Statement
A cubic cardboard box of side a = 0.250 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= 3.20 N/C.m is a constant.
http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype53/prob07a_cube.gif
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.)
Hint:Since E is not constant, you will need to do an area integral for this problem. Remember the definition of electric flux. Only one component of E contributes to the flux out of the top of the box. In doing the integral to find the flux, it might help to imagine breaking the top surface into strips along which the z-component of the electric field is constant. If all else fails, ask for help.
Homework Equations
Flux=[tex]\int[/tex]E*dA
The Attempt at a Solution
The Nonuniform part is what's tripping me up. I know how to do it with uniform electric field.
I'll call the top face faceA, Flux=I. So I[tex]_{A}[/tex]=[tex]\int[/tex]E*dA= [tex]\int[/tex](3.20 N/C z [tex]\widehat{j}[/tex] + 3.20 N/C y[tex]\widehat{k}[/tex])* (dA [tex]\widehat{k}[/tex])
So it says that for this face Z=the side length, but what do I do with y? and with the top face wouldn't only the K-hat vectors apply anyway?
and once I get an asnwer will it be negative?
anyway if someone could give me some insight, let me know if I've set it up right it would help a lot. Thanks in advance!
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