Force on a particle of a linear charge distribution

AI Thread Summary
The discussion centers on calculating the force exerted by a linear charge distribution on a point charge. The user encounters a sign issue in their electric field calculation, specifically in the evaluation of an integral related to the electric field. They derive the electric field expression but question the correctness of their antiderivative, which leads to a negative force result. A response points out that the user should verify whether the derivative of their antiderivative matches the original integrand, indicating a potential error in their integration process. The discussion emphasizes the importance of careful evaluation of integrals in electrostatics calculations.
Guillem_dlc
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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.
Captura de 2022-03-20 16-08-22.png


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

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}##?
 
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