# Line integral ad electric charge

• Kokordilos
In summary, the homework statement says that the force exerted by an electric charge at the origin on a charged particle at point (x,y,z) with position vector r = <x,y,z> is F(r) = Kr / |r|^3. To find the work done as the particle moves along a straight line from (2,0,0) to (2,1,5), you need to find the force function F(r) = Kr / |r|^3 and replace the x, y, z in the force vector by (2, t, 5t), and integrate from t= 0 to t= 1.
Kokordilos

## Homework Statement

The force exerted by an electric charge at the origin on a charged particle at point (x,y,z) with position vector r = <x,y,z> is F(r) = Kr / |r|^3
where K is a constant. Find the work done as the particle moves along a straight line from (2,0,0) to (2,1,5).

## The Attempt at a Solution

So if it moves from (2,0,0) to (2,1,5) I can parametrize:
x = 2
Y = T
Z = 5T
for 0<t<1

so r(t) = (2, T, 5T)
r'(t) = (0, 1, 5)
But how do I put the force function F(r) = Kr / |r|^3 in terms of these parameters so I can do the integral?

The electrostatic force is a conservative force. This means that the work done by the electrostatic force while moving a charge from one point to another is independent of the path and only depends on the end points. The magnitude of the work done by the electrostatic force will be the difference in the potential energy of the system in the two configurations. Using this you can find the answer to your question without computing the line integral along the path requested, but by choosing another simpler path.

However, if that is not allowed, then you probably should re-express the force law you were given in terms of Cartesian coordinates.

work= force* distance moved and if the force is variable along the path, it is \int f dx[/itex]. Finally if force is a vector, then it is $\vec{F}\dot\vec{ds}$ where the integral is take along the path the object moves on.

Here the force is given as $$\frac{K\vec{r}}{|\vec{r}|^3}[/itex] which, in Cartesian coordinates, is [tex]\frac{K}{(x^2+ y^2+ z^2)^{3/2}}(x\vec{i}+ y\vec{j}+ z\vec{k}$$. As you say, parametric equations for the path are x= 2, y= t, z= 5t so you can replace x, y, z in the force vector by those. As a vector, the path is given by $$\vec{r}= 2\vec{i}+ t\vec{j}+ 5t\vec{j}$$ so [itex]d\vec{r}= (\vec{j}+ 5\vec{k})dt[/tex]. Integrate the dot product of those two from t= 0 to t= 1.

However, AEM is correct. That is a conserative force so you could also find a "potential", an antiderivative, F(x,y,z) such that $$\nabla F$$ is the force vector, and evaluate that at the two end points. You would have to find F such that
$$\frac{\partial F}{\partial x}= \frac{Kx}{(x^2+ y^2+ z^2)^{3/2}}$$
$$\frac{\partial F}{\partial y}= \frac{Ky}{(x^2+ y^2+ z^2)^{3/2}}$$
$$\frac{\partial F}{\partial z}= \frac{Kz}{(x^2+ y^2+ z^2)^{3/2}}$$

I'm very happy to read your post, because that's actually how i ended up doing it, though I certainly wasn't sure if it was right. Thanks very much, I really appreciate all your help!

## 1. What is a line integral and how is it related to electric charge?

A line integral is a mathematical concept used to calculate the total value of a function along a specific path. In the case of electric charge, the line integral is used to calculate the total electric field along a given path. By integrating the electric field over a path, we can determine the work required to move a unit charge along that path.

## 2. How is the line integral of electric charge calculated?

The line integral of electric charge is calculated by multiplying the electric field along a path by the differential length of that path and then summing up these values. This can be represented mathematically as ∫C E•dl, where C is the path of integration, E is the electric field, and dl is the differential length along the path.

## 3. What is the purpose of calculating a line integral of electric charge?

The purpose of calculating a line integral of electric charge is to determine the total amount of work required to move a unit charge along a specific path in an electric field. This can be useful in various applications, such as designing circuits and understanding the behavior of electric currents.

## 4. How does the direction of the path affect the line integral of electric charge?

The direction of the path does affect the line integral of electric charge. This is because the electric field is a vector quantity and its direction can change along different paths. Therefore, the line integral will be different for different paths, depending on the direction of the electric field along those paths.

## 5. Can the line integral of electric charge be negative?

Yes, the line integral of electric charge can be negative. This can happen when the electric field and the path of integration are in opposite directions. In this case, the work required to move a unit charge along the path is negative, indicating that the electric field is doing work on the charge and not the other way around.

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