Electric field created by two charged circular arcs?

In summary, the strategy will be to figure out what ##dq##, ##\hat{r}_{dq,p}##, and ##r_{dq,p}## are, plug them into the expression for ##d\vec{E}_{p_r}##, then integrate over ##d\vec{E}_{p_r}## to obtain ##\vec{E}_{p_r}##, the electric field at ##P## due to the arc on the right.Then I will repeat the process to calculate ##d\vec{E}_{p_l}## and ##\vec{E}_{p_l}##, for the arc on the left. The
  • #1
zenterix
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Homework Statement
Two circular arcs of radius ##R## are uniformly charged with a positive charge per unit length ##\lambda##. The arcs lie on a plane as shown in the figure below. Each arc subtends an angle ##\theta=\frac{\pi}{3}##.

What is the direction and magnitude of the electric field anywhere along the ##z## axis that passes through the center of the circular arcs, perpendicular to the plane of the figure?
Relevant Equations
Infinitesimal electric field created by an infinitesimal charge ##dq## on the right-side arc
$$d\vec{E}_{p_r}=\frac{k_e dq}{r_{dq,p}^2}\hat{r}_{dq,p}$$
The strategy will be to figure out what ##dq##, ##\hat{r}_{dq,p}##, and ##r_{dq,p}## are, plug them into the expression for ##d\vec{E}_{p_r}##, then integrate over ##d\vec{E}_{p_r}## to obtain ##\vec{E}_{p_r}##, the electric field at ##P## due to the arc on the right.

Then I will repeat the process to calculate ##d\vec{E}_{p_l}## and ##\vec{E}_{p_l}##, for the arc on the left. The latter will be basically the same result as for ##d\vec{E}_{p_r}## but with one sign changed.

$$dq=\lambda ds = \lambda R d\theta$$

Here are the original sketch of the problem and my own sketch
1636711870367.png
1636711928531.jpeg

$$\vec{r}_{dq,p}=\vec{r}_{0,p}-\vec{r}_{0,dq}$$

$$\vec{r}_{0,p}=z\hat{k}$$

$$\hat{r}_{0,dq}=\cos{\theta}\hat{i} +\sin{\theta}\hat{j}$$

$$\vec{r}_{0,dq}=R\hat{r}_{0,dq}=R(\cos{\theta}\hat{i} +\sin{\theta}\hat{j})$$

$$\implies \vec{r}_{dq,p}=z\hat{k}-R\cos{\theta}\hat{i} -R\sin{\theta}\hat{j}$$

$$d\vec{E}_{p_r}=\frac{k_e \lambda R d\theta}{(z^2+R^2)^{3/2}}(z\hat{k}-R\cos{\theta}\hat{i} -R\sin{\theta}\hat{j})$$

The expression for ##d\vec{E}_{p_l}## is derived analogously, and the only thing that changes is s sign on ##\hat{r}_{0,dq}##

$$\hat{r}_{0,dq}=-\cos{\theta}\hat{i} +\sin{\theta}\hat{j}$$

$$d\vec{E}_{p_l}=\frac{k_e \lambda R d\theta}{(z^2+R^2)^{3/2}}(z\hat{k}+R\cos{\theta}\hat{i} -R\sin{\theta}\hat{j})$$

Now we integrate over ##\theta##

$$\vec{E}_{p_r}=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{k_e \lambda R d\theta}{(z^2+R^2)^{3/2}}(z\hat{k}-R\cos{\theta}\hat{i} -R\sin{\theta}\hat{j})$$

$$\vec{E}_{p_l}=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{k_e \lambda R d\theta}{(z^2+R^2)^{3/2}}(z\hat{k}+R\cos{\theta}\hat{i} -R\sin{\theta}\hat{j})$$

We end up with

$$\vec{E}_{p_r}=\frac{k_e \lambda R d\theta}{(z^2+R^2)^{3/2}}(\frac{2\pi z}{3}\hat{z} -R\sqrt{3}\hat{i})$$$$\vec{E}_{p_l}=\frac{k_e \lambda R d\theta}{(z^2+R^2)^{3/2}}(\frac{2\pi z}{3}\hat{z} +R\sqrt{3}\hat{i})$$

Therefore

$$\vec{E}_p=\vec{E}_{p_r}+\vec{E}_{p_l}=\frac{k_e \lambda R d\theta}{(z^2+R^2)^{3/2}}\frac{4\pi z}{3}\hat{k}$$
$$=\frac{\lambda R z}{3\epsilon_0 (z^2+R^2)^{3/2}}\hat{k}$$

I'd like to know if this solution is correct because I am following along on MIT Open Learning Library and this is a practice problem. I am allowed to submit answers, and am given a correct/incorrect feedback. For the expression above, I am getting incorrect, though I can't figure out why.
 
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  • #2
Hi @zenterix. Your integration limits should be ##-\frac {\pi}{6}## to ##+\frac {\pi}{6}## (to cover a total angle of ##\frac {\pi}{3}## for each arc).

Are you required to do the formal vector/integration method as an exercise? The problem can be solved in a few lines of basic algebra using some simple observations:
- by symmetry, the x and y components of the field at P are zero;
- the magnitude of the field at P from each charge-element (dq) is ##dE = \frac {kdq}{z^2 + R^2}##;
- the z-component of ##\vec {dE}## is ##dE cos(\phi)## where ##\phi## is the angle between the z-axis and the line from P to the charge-element.

It's a useful exercise to try as you will see how much easier/quicker the use of symmetry can make a problem.
 
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  • #3
I can't believe I missed the integral limits! My god.

I am aware of the symmetry of the problem. At this stage I am doing the full calculations to get practice doing the calculus. But I will try to just use symmetry to also get practice with identifying such shortcuts from now on as well.
 
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Related to Electric field created by two charged circular arcs?

1. What is an electric field created by two charged circular arcs?

The electric field created by two charged circular arcs is a type of electric field that is formed when two circular arcs with opposite charges are placed near each other. This type of electric field is also known as a dipole field.

2. How is the electric field strength calculated for two charged circular arcs?

The electric field strength for two charged circular arcs can be calculated using the equation E = k*(q1/r1^2 - q2/r2^2), where k is the Coulomb's constant, q1 and q2 are the charges of the two arcs, and r1 and r2 are the distances from the center of each arc to the point where the electric field is being measured.

3. What factors affect the strength of the electric field created by two charged circular arcs?

The strength of the electric field created by two charged circular arcs is affected by the magnitude of the charges on the two arcs, the distance between the two arcs, and the angle between the two arcs.

4. How does the direction of the electric field change between the two charged circular arcs?

The direction of the electric field between the two charged circular arcs changes from positive to negative as you move from one arc to the other. This means that the electric field lines will point away from the positive arc and towards the negative arc.

5. What are some real-life applications of the electric field created by two charged circular arcs?

The electric field created by two charged circular arcs has many practical applications, such as in capacitors, particle accelerators, and electronic devices. It is also used in medical imaging techniques, such as magnetic resonance imaging (MRI), to create images of the human body.

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