Force on a particle of a linear charge distribution

[Guillem_dlc]

Hello!

I am trying to solve this exercise of the electric field, but it comes out changed sign and I don't know why.

Statement: On a straight line of length $L=60\, \textrm{cm}$ a charge $Q=3,0\, \mu \textrm{C}$ is uniformly distributed. Calculate the force this linear distribution makes on a point charge $q=5,0\, \mu \textrm{C}$ in the same direction of the thread and at a distance of $30\, \textrm{cm}$ from one of its ends.

My solution: First, we want to look at how much the electric field is worth at different points. We choose an infinitesimal charge $dq$ any at any point in the thread $(x,0)$, and we assign to the point charge any $(x_0,0)$ that fulfils the condition $x_0>L$. According to this, we can already define the vector $\overrightarrow{r}$ that arises in the charge differential and ends at point $(x_0,0)$:
$$\overrightarrow{r}=(x_0,0)-(x,0)=(x_0-x,0)$$
$$r=x_0-x\rightarrow \widehat{r}=\dfrac{\overrightarrow{r}}{r}=(1,0)$$
$$\lambda =\dfrac{dq}{dl}\rightarrow dq=\lambda dl$$
where $dl=dx$ because we only have component $x$.
$$E=\int_L k\dfrac{dq}{r^2}\widehat{r}=k\int_0^L \dfrac{\lambda \, dx}{(x_0-x)^2}(1,0)$$
Then,
$$E=k\lambda \widehat{i}\int_0^L \dfrac{1}{(x_0-x)^2}\, dx = k\lambda \widehat{i} \left[ \boxed{\dfrac{-1}{(x_0-x)}}\right]_0^L = k\lambda \widehat{i} \left( \dfrac{-1}{x_0-L}+\dfrac{1}{x_0}\right)$$
$$=\dfrac{-k\lambda}{x_0-L}+\dfrac{k\lambda}{x_0}=\dfrac{-k\lambda}{0,3}+\dfrac{k\lambda}{0,9}=\dfrac{-3k\lambda +k\lambda}{0,9}=$$
$$=\dfrac{-2k\lambda}{0,9}=-100000\, \textrm{V}/\textrm{m},$$
using that $\lambda =\dfrac{dq}{dL}=\dfrac{Q}{L}=0,000005$. Finally,
$$F=qE=-0,5\, \textrm{N}$$

My question: It gives me good, but it's changed in sign and I think that's why I've marked in the integral, but I'd say I've done it well...

 

[TSny]

$$E=k\lambda \widehat{i}\int_0^L \dfrac{1}{(x_0-x)^2}\, dx = k\lambda \widehat{i} \left[ \boxed{\dfrac{-1}{(x_0-x)}}\right]_0^L$$

You are right to suspect that the problem is with the evaluation of the integral. You got $$\dfrac{-1}{(x_0-x)}$$ for an antiderivative of $$\dfrac{1}{(x_0-x)^2}$$

Does the derivative of $\dfrac{-1}{(x_0-x)}$ with respect to $x$ equal $\dfrac{1}{(x_0-x)^2}$?