- #1
danago
Gold Member
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A point particle with charge q is placed at a corner of a cube of edge a . What is the flux through (a) each cube face forming that corner and (b) each of the other cube faces?
I tried modelling the problem by positioning the charge such that it is at point (0,0,0) in the x-y-z space. The cube then extends out in the positive x, y and z directions.
I then defined the unit vectors i, j and k to extend in the x, y and z directions respectively. Now I am not sure if my next step is valid, but it seemed to be a decent thing to do. I said that the electric field due to the charge is given by:
<br /> \overrightarrow E = \frac{{kq}}{{x^2 }}\widehat{\underline i } + \frac{{kq}}{{y^2 }}\widehat{\underline j } + \frac{{kq}}{{z^2 }}\widehat{\underline k }<br />
I then evaluated the integral of the dot product of vectors E and dA over one of the cubes faces, and came up with the expression for flux as kq, however this wasnt the correct result.
Anyone able to help with the question?
Thanks,
Dan.
I tried modelling the problem by positioning the charge such that it is at point (0,0,0) in the x-y-z space. The cube then extends out in the positive x, y and z directions.
I then defined the unit vectors i, j and k to extend in the x, y and z directions respectively. Now I am not sure if my next step is valid, but it seemed to be a decent thing to do. I said that the electric field due to the charge is given by:
<br /> \overrightarrow E = \frac{{kq}}{{x^2 }}\widehat{\underline i } + \frac{{kq}}{{y^2 }}\widehat{\underline j } + \frac{{kq}}{{z^2 }}\widehat{\underline k }<br />
I then evaluated the integral of the dot product of vectors E and dA over one of the cubes faces, and came up with the expression for flux as kq, however this wasnt the correct result.
Anyone able to help with the question?
Thanks,
Dan.
That made it much easier than whatever i was trying to do originally haha.
) \Phi=q / \epsilon_0 is the total flux in all directions. How much goes through that cube?
