# What Is the Electric Field of a Curved Rod at the Origin?

• Zack K
In summary: You have to take the limit. Doing so should give you Coulomb’s law for a point charge at (x,y) = (R,0) because that is the situation α=0 describes. (But again, you’ve only determined the field at a particular... location!)
Zack K

## Homework Statement

A rod of charged -Q is curved from the x-axis to angle ##\alpha##. The rod is a distance R from the origin (I will have a picture uploaded). What is the electric field of the charge in terms of it's x and y components at the origin? k is ##\frac {1} {4\pi \epsilon_0}##

## Homework Equations

##\vec E=\frac {kQ} {r^2}##

## The Attempt at a Solution

Let ##\theta## represent a any angle of ##\alpha##
##\vec r=\langle -Rcos\theta, -Rsin\theta, 0\rangle##
##|\vec r|=\sqrt {(-Rcos\theta)^2+(-Rsin\theta)^2}##

$$\hat r=\frac {\vec r} {\vec |r|}=\frac {\langle-cos\theta, -sin\theta, 0\rangle} {\sqrt{-cos\theta)^2+(-sin\theta)^2}}=\frac {\langle-cos\theta, -sin\theta, 0\rangle}{1}$$You can factor out the R from the top and bottom and cancel them out. The denominator just becomes 1

Now representing ##\Delta Q##
$$\Delta Q=Q\frac{\Delta arclength} {arclength}=Q\frac{R\Delta\theta} {R\alpha}=Q\frac{\Delta\theta} {\alpha}$$
The R's will cancel out.
Now assuming rod consists of many point charges
$$\Delta {\vec E}=\frac {k\Delta Q}{|\vec r|^2} \hat r=\frac {kQ\Delta \theta}{R^2\alpha}\langle-cos\theta, -sin\theta, 0\rangle$$Now we integrate in terms of ##\theta## and integrate each component. $$\Delta {\vec E_x}=\int_{0}^{\alpha} \frac {kQ(-cos\theta)}{R^2\alpha}d\theta=\frac {kQ}{R^2}\int_{0}^{\alpha} \frac {(-cos\theta)}{\alpha}d\theta$$ $$\Delta {\vec E_y}=\int_{0}^{\alpha} \frac {kQ(-sin\theta)}{R^2\alpha}d\theta=\frac {kQ}{R^2}\int_{0}^{\alpha} \frac {(-sin\theta)}{\alpha}d\theta$$This is where I'm stuck on. Not entirely sure if I can factor an alpha out.

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Zack K said:
This is where I'm stuck on. Not entirely sure if I can factor an alpha out.
Why not? You are integrating over ##\theta## not ##\alpha## which is a given of the problem and which you may assume is constant. This will allow you to find the x and y components as functions of ##\alpha##. You can test your work by seeing if you get the expected direction for ##\alpha = \pi/2, \pi, 3\pi/2##.

kuruman said:
Why not? It is a given of the problem so you may assume it is constant. This will allow you to find the x and y components as functions of ##\alpha##. You can test your work by seeing if you get the expected result for ##\alpha = \pi/2, \pi, 3\pi/2##.
The problem I see is that for the x component when I integrate -cos it becomes -sin. But then if I put -sin(0) I would get 0 electric field on the x-axis which doesn't make sense.

Zack K said:
The problem I see is that for the x component when I integrate -cos it becomes -sin. But then if I put -sin(0) I would get 0 electric field on the x-axis which doesn't make sense.
When you integrate the trig function, you get ##-\sin \theta |_0^{\alpha}##. What does that give you?

kuruman said:
When you integrate the trig function, you get ##-\sin \theta |_0^{\alpha}##. What does that give you?
##-sin\alpha## but, wouldn't this mean that I have no variables in my function then? All I would have are constants.

Zack K said:
##-sin\alpha## but, wouldn't this mean that I have no variables in my function then? All I would have are constants.
Your variables are ##Q##, ##\alpha## and ##R##. You are calculating ##E_x## and ##E_y## if ##Q##, ##\alpha## and ##R## are given to you. What else is there to be said?

kuruman said:
Your variables are ##Q##, ##\alpha## and ##R##. You are calculating ##E_x## and ##E_y## if ##Q##, ##\alpha## and ##R## are given to you. What else is there to be said?
But the equation let's say for ##\vec E_x=\frac {kQ(-sin\alpha)}{R^2\alpha}## doesn't seem like it holds. I don't see any variables, alpha is a constant. Even if I treated it as a variable, and wanted to find ##\vec E_x## at ##\alpha=0## Then I would get sin(0) which is 0 and isn't true in this case(I uploaded a diagram). Not to mention how you would have a 0 on the denominator.

Zack K said:

## Homework Equations

##\vec E=\frac {kQ} {r^2}##
This is probably just a typo, but you cannot equate a (multi-dimensional) vector with a scalar
Zack K said:
Now assuming rod consists of many point charges
$$\Delta {\vec E}=\frac {k\Delta Q}{|\vec r|^2} \hat r=\frac {kQ\Delta \theta}{R^2\alpha}\langle-cos\theta, -sin\theta, 0\rangle$$
Remember, the charge is given as negative Q.
Zack K said:
But the equation let's say for ##\vec E_x=\frac {kQ(-sin\alpha)}{R^2\alpha}## doesn't seem like it holds. I don't see any variables, alpha is a constant.
Perhaps you’re wondering why the field does not depend on x,y position? Remember, you’ve only calculated the field at a particular point! (At the origin.)
Zack K said:
Even if I treated it as a variable, and wanted to find ##\vec E_x## at ##\alpha=0## Then I would get sin(0) which is 0 and isn't true in this case(I uploaded a diagram). Not to mention how you would have a 0 on the denominator.
0/0 is not necessarily zero. You have to take the limit. Doing so should give you Coulomb’s law for a point charge at (x,y) = (R,0) because that is the situation α=0 describes. (But again, you’ve only determined the field at a particular point.)

Zack K
Everything I had to say has already been covered by @Hiero in the preceding post. Just think about it.

Zack K
Hiero said:
This is probably just a typo, but you cannot equate a (multi-dimensional) vector with a scalar
Yes sorry that was a typo.
Hiero said:
Remember, the charge is given as negative Q.
That was a typo as well.
Hiero said:
Perhaps you’re wondering why the field does not depend on x,y position? Remember, you’ve only calculated the field at a particular point! (At the origin.)
Ah I see I just realized that the equation is for the origin and is a constant value at the origin. There would be no reason to have a variable since the origin is fixed.

@kuruman
@Hiero
Thank you guys so much.

## 1. What is the definition of an electric field?

The electric field is a physical field that surrounds electrically charged particles and exerts a force on other charged particles within the field. It is a fundamental concept in electromagnetism and is measured in units of volts per meter (V/m).

## 2. How is the electric field of a curved rod different from a straight rod?

The electric field of a curved rod is different from a straight rod because of the curvature of the rod. The electric field lines are closer together at the curved parts of the rod, resulting in a stronger electric field. Additionally, the electric field may be more concentrated at the tips of the curved rod.

## 3. What factors affect the strength of the electric field of a curved rod?

The strength of the electric field of a curved rod is affected by the magnitude and distribution of charge on the rod, as well as the curvature of the rod itself. The distance from the rod also plays a role, as the electric field decreases with distance.

## 4. Can the electric field of a curved rod be calculated?

Yes, the electric field of a curved rod can be calculated using the principles of electrostatics. The electric field can be calculated at any point in space by considering the distance from the rod, the magnitude of charge on the rod, and the curvature of the rod. Numerical methods or calculus may be necessary for more complex curved rods.

## 5. How is the electric field of a curved rod used in practical applications?

The electric field of a curved rod can be used in various applications, such as electrostatic painting, particle accelerators, and electronic devices. It is also a fundamental concept in understanding the behavior of charged particles in electric fields and is used in fields such as physics, engineering, and chemistry.

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