What is the electric flux through the top surface of the cube?

  • Thread starter jincy34
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  • #1
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Homework Statement



A 3 nC point charge is at the center of 5 m x 5 m x 5 m cube. What is the electric flux through the top surface of the cube?

Homework Equations



E=kq/r^2
φ = E.A

The Attempt at a Solution



E = (8.99*109)(3*10-9) / 2.52 = 4.3152 N/C

φ = E.A = (4.3152)(5*5) = 107.88
But, it is wrong. Can someone please tell me what I am doing wrong?
 

Answers and Replies

  • #2
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You assume a constant electric field over the whole top surface

The equation E=kq/r^2 where r is, I assume you put in 5, only the electric field at the center of the top. Every other point on the top is gonna have a different r and electric field, it's not constant.

Edit: actually for r it looks like you used...sqrt(2.52)? I'm not sure where that came from
 
Last edited:

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