Electric field integration? [closed]

If we have a rod uniformly charged with $Q$ stretching from $-a$ to $a$ on the $x$-axis as shown in the picture

And we want to calculate electric field in point $2a$ on the $x$-axis, we know that the electric field of a point charge $dQ$ on the rod in point $2a$ is: $$\vec{dE}=\frac{dQ}{4\pi\epsilon_0r^2}\vec{i_x}$$ where $r$ is the distance between the charge and the point $2a$.

We know that $dQ=Q'dl$ and, in this case, $Q'=\frac{Q}{2a}$, so $dQ=\frac{Q}{2a}dl$ and finally $$\vec{dE}=\frac{Qdl}{8a\pi\epsilon_0r^2}\vec{i_x}$$

Now, to calculate the total field, we need to sum all the fields of all point charges on the charged rod in point $2a$ and we do that by integration. $$E=\int \limits_{a}^{3a}dE=\frac{Q}{8a\pi\epsilon_0}\int \limits_{a}^{3a}\frac{dr}{r^2}=\frac{Q}{8a\pi\epsilon_0}\bigg(\frac{1}{a}-\frac{1}{3a}\bigg)=\frac{Q}{12\pi\epsilon_0a^2}$$

That is where I get confused. Why can't we integrate from $3a$ to $a$? We would get the same field intensity, but with a minus sign, which would be correct for the point $-2a$, or would we? I think I'm missing something here regarding the integration direction and the vectors.

closed as off-topic by user36790, Danu, Kyle Kanos, Martin, Sebastian RieseFeb 18 '16 at 21:40

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Your last statement is wrong since interchanging the limits of the integrals will change the sign of the integral. Anyway I will try to derive the same expression by taking a general case and show you that the sign of the x-coordinate doesn't matter rather it purely depends on the distance of point from the center of the rod.

Lets take dx rather than dl where x is the x coordinate of the element dx (you called it dl).

dx is an element of the rod which whose x-coordinate is x. Since r is the distance of the dx element on the rod to the point (2a, 0), r can be substituted as (2a - x). Notice that when x becomes negative, (2a - x) becomes (2a + |x|) , i.e: the distance increases correctly.

Now using your differential equation and substituting the variables correctly, we get

$dE = \frac {Qdx}{8a\pi\epsilon r^2}$

Substituting for r but instead of using 2a, lets use d and solve for a general case.

$dE = \frac {Qdx}{8a\pi\epsilon (d - x)^2}$

Now integrating both sides,

$E = \frac {Q}{8a\pi\epsilon}\int_{-a}^a \frac {1}{(d - x)^2} = \frac {Q}{8a\pi\epsilon} (\frac {1}{d - a} - \frac {1}{d + a}) = \frac {Q}{8a\pi\epsilon} \frac {2a}{d^2 - a^2} = \frac {Q}{4a\pi\epsilon} \frac {a}{d^2 - a^2}$

Notice that we have $d^2$ in the equation which answers your doubt. You don't get a minus when you substitute -2a which is in agreement with the predictions from symmetry. (The nature wouldn't cheat between left and right, hence one should expect the magnitude to be same and the field to be either away from the rod or towards the rod weather or not you move distance d to the left or right).

Now lets substitute d as 2a, you shall get the magnitude to be,

$E = \frac {Q}{12\pi\epsilon a^2}$

• A6SE: To emphasize the point made here, try to stay with the coordinate system you have chosen and express distances in terms of the coordinates rather than substituting mere distances like you did. In the long run, it will save you from confusion when the charge distributions get more complicated. – Bill N Feb 16 '16 at 18:00