- #1
jegues
- 1,097
- 3
DiracRules said:First of all, you did not write correctly the expression of the electric field in Cartesian coordinates. (Remember that |r| is not -usually- equals to 1)
Then, how do you usually evaluate the surface integral (the flux) over a certain surface?
DiracRules said:Sorry for misunderstanding r, but everyone has his own conventions and symbols :D
In Cartesian coordinates, how do you write the position occupied by an object? By giving the three coordinates. You can write both [itex] \vec{r}_P=\left[\begin{array}{c}x\\y\\z\end{array}\right][/itex] and [itex]\vec{r}_P=x\hat{i}+y\hat{j}+z\hat{k}[/itex] since[itex]\left[\begin{array}{c}x\\y\\z\end{array}\right]=x\left[\begin{array}{c}1\\0\\0\end{array}\right]+y\left[\begin{array}{c}0\\1\\0\end{array}\right]+z\left[\begin{array}{c}0\\0\\1\end{array}\right][/itex]
Was this your problem?
Now, I think that for the first part of the question you cannot use the flux theorem because the problem asks to calculate the flux through one face of the cube, not through the whole surface: I fear you need to calculate explicitly the flux by evaluating the surface integral or something like that (it shouldn't be too difficult, however).
You'd better express the electric field in term of its cartesian components and then try to integrate.
DiracRules said:I think it is right.
To evaluate the integral, you can use http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions" > List of integrals involving [itex]R=\sqrt{ax^2+bx+c}[/itex] > [itex]\int \frac{dx}{R^3}[/itex]