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Let us consider a positive charge placed at a distance $x$ on left outside the cube of side $a$. The electric fields are entering the cube from left and exiting from the right. Will the flux in the cube be zero or will it have a finite value?

The electric field varies inversely with distance as $E$ proportional to $1/r^2$. Since area is same, would the flux through the left side be greater than flux through right?

$$E.A=A.kq/x^2$$ $$E'.A=A.kq/(x+a)^2$$

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Hint: Apply guass law

The mistake you are doing is that the field E=kq/r^2 is valid at a distance r from the point charge(so spherical symmetry). All points on the face of the cube are not at a distance of x or x+a from the point charge.

Also as garyp mentioned in the comment below, all field lines do not exit from the opposite faces, so finding flux through all faces by considering strips and integrating would be a nightmare.

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  • $\begingroup$ In addition, not all electric field lines exit from the opposite face. BTW, I think (my opinion) you give away too much by answering the question in your first sentence. $\endgroup$
    – garyp
    Commented Sep 22, 2019 at 17:48
  • $\begingroup$ Oh sorry I changed my first sentence now. $\endgroup$
    – user600016
    Commented Sep 23, 2019 at 1:13

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