Find the electric field of a charged arc a distance R away

[Jaccobtw]

Homework Statement

find electric field of a quarter circle charged arc at point located the same distance from all points on the arc

Relevant Equations

$$dQ = \lambda R d \theta$$
$$dE_x = \frac{k_e dQ}{R^2} cos \theta$$

define charge at an infinitesimal length of arc

$$dQ = \lambda R d \theta$$We only care about the x component of the electric field because the y components cancel due to symmetry

$$dE_x = \frac{k_e dQ}{R^2} cos \theta$$

Integrate to add up the infinitesimal parts. A quarter circle means 90 degrees so integrate from 0 to pi/2.$$\int dE_x=\int_{0}^{\frac{\pi}{2}} \frac{k_e \lambda R d \theta}{R^2} cos \theta$$

PROBLEM: I get two different answers when I integrate from 0 to pi/2 and -pi/4 to pi/4. The right answer came from when I used -pi/4 to pi/4.

Doe anyone know why you can't integrate from 0 to pi/2?

Thanks

 

[TSny]

PROBLEM: I get two different answers when I integrate from 0 to pi/2 and -pi/4 to pi/4. The right answer came from when I used -pi/4 to pi/4.

Doe anyone know why you can't integrate from 0 to pi/2?

Thanks

If the quarter circle goes from $\theta = 0$ to $\theta = \pi/2$, would there be a net y-component of the electric field as well as a net x-component of the field at the point where you are calculating the field? I'm assuming $\theta = 0$ corresponds to the positive x-axis.

 

[Jaccobtw]

If the quarter circle subtends the angle between $\theta = 0$ amd $If the quarter circle goes from$\theta = 0$to$\theta = \pi/2$, would there be a net y-component of the electric field as well as a net x-component of the field at the point where you are calculating the field? I'm assuming$\theta = 0## corresponds to the positive x-axis.

I think this makes sense. Do you mind explaining why I also get the right answer if i integrate from 0 to pi/4 twice?

 

[TSny]

Do you mind explaining why I also get the right answer if i integrate from 0 to pi/4 twice?

Can you see why integrating from $-\pi/4$ to $0$ contributes the same amount to $E_x$ as integrating from $0$ to $\pi/4$?

 

[Jaccobtw]

Can you see why integrating from $-\pi/4$ to $0$ contributes the same amount to $E_x$ as integrating from $0$ to $\pi/4$?

Never mind. I think I get it. How could you set up the integral if you wanted to use pi/2 to 0

 

[TSny]

How could you set up the integral if you wanted to use pi/2 to 0

If the quarter circle extends from the x-axis to the y-axis, so that $\theta$ varies from $0$ to $\pi/2$, then the integral that you set up will give you $E_x$ when integrated from $\theta = 0$ to $\theta = \pi/2$. (But think about the sign of $E_x$.)

How would you set up the integral to get $E_y$ for this case? How would you combine $E_x$ and $E_y$ to get the total electric field?

Why didn't you need to worry about $E_y$ when you integrated from $-\pi/4$ to $\pi/4$?

 

[Jaccobtw]

Why didn't you need to worry about $E_y$ when you integrated from $-\pi/4$ to $\pi/4$?

Because the y components cancel. I'm confused about how I'd set it up for 0 to pi/2 though to get the magnitude of the electric field. I thought it would be $cos^2 \theta + sin^2 \theta$ but and I got an answer that's kinda close but still off. Any ideas?

 

[TSny]

Because the y components cancel.

Yes, for the case of $-\pi/4$ to $\pi/4$, the net y component is zero

I'm confused about how I'd set it up for 0 to pi/2 though to get the magnitude of the electric field. I thought it would be $cos^2 \theta + sin^2 \theta$ but and I got an answer that's kinda close but still off. Any ideas?

For this case, what do you get when you evaluate the integral for $E_x$? Similarly, what do you get for $E_y$?

Neither of these components will have $\theta$ in the answer. ($\theta$ was the integration variable and the integrals have been done.) So, when you combine $E_x$ and $E_y$ to get the magnitude of the total electric field vector, $\cos^2 \theta + \sin^2 \theta$ will not come into play.

 

[Jaccobtw]

Yes, for the case of $-\pi/4$ to $\pi/4$, the net y component is zeroView attachment 298399
For this case, what do you get when you evaluate the integral for $E_x$? Similarly, what do you get for $E_y$?

Neither of these components will have $\theta$ in the answer. ($\theta$ was the integration variable and the integrals have been done.) So, when you combine $E_x$ and $E_y$ to get the magnitude of the total electric field vector, $\cos^2 \theta + \sin^2 \theta$ will not come into play.

I honestly have no idea

 

[TSny]

You said in your first post that you worked out $E_x$ for the case where $\theta$ ranges from $0$ to $\pi/2$. But you didn't state what you got for the answer for $E_x$. What did you get?

Setting up an integral for $E_y$ should be very similar. Can you show the integral that you would evaluate to obtain $E_y$?

Alternately, using symmetry you should be able to see how $E_y$ compares to $E_x$ without actually carrying out the integration for $E_y$.