Force on a particle of a linear charge distribution

In summary, the conversation is about a problem with the evaluation of an integral in an exercise on electric fields. The expert suggests that the issue might be with the antiderivative used and asks if the derivative of the antiderivative matches the given function.
  • #1
Guillem_dlc
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Thread moved from the technical forums to the schoolwork forums
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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  • #2
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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Related to Force on a particle of a linear charge distribution

1. What is the definition of "force on a particle of a linear charge distribution?"

The force on a particle of a linear charge distribution is the electrostatic force exerted on the particle due to the presence of other charges in the distribution.

2. How is the force on a particle of a linear charge distribution calculated?

The force on a particle of a linear charge distribution can be calculated using Coulomb's Law, which states that the force is equal to the product of the two charges divided by the square of the distance between them.

3. Can the force on a particle of a linear charge distribution be repulsive?

Yes, the force on a particle of a linear charge distribution can be either attractive or repulsive, depending on the signs of the charges involved. Like charges will repel each other, while opposite charges will attract.

4. How does the distance between the particle and the linear charge distribution affect the force?

The force on a particle of a linear charge distribution is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases.

5. What are some real-life applications of the force on a particle of a linear charge distribution?

The force on a particle of a linear charge distribution is involved in many everyday phenomena, such as the attraction between a comb and hair, the operation of electric motors, and the behavior of lightning strikes.

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