# Electric Field due to a charged line; With y as function of theta

• Molderish
In summary, the main issue in this conversation is how to express "y" as a function of theta in order to solve for the integral. The conversation discusses using two methods - one being to make the substitution of cosine(theta) = x/sqrt(x^2+y^2) and then integrating, and the other being to use the derivative of y with respect to theta and then multiplying both sides by dtheta. The latter method is demonstrated with an example using LaTeX. Ultimately, the goal is to find a solution for the integral of cos(theta)dy/(x^2+y^2) with appropriate limits.
Molderish
Well the main problem here its once a i get the integral i do not how to put dy as function of thetha and then integrate ; at the end that's what i get

Ex=integral (dE*cos$\theta$)=($\lambda$/4$\pi$$\epsilon$)
times the integral of cos($\theta$)dy/x^2+y^2

where i can easily find the cosine and do the integral but I'm asked to express strictly "y" as function of thetha so as cos(θ)=x/√x^2+y^2 so 1/√x^2+y^2=cos^2(θ)/x^2

and i know that y=x*tan(θ) but i don't how to get "dy" graphically nor analytically;

#### Attachments

• IMG_20140216_194726.jpg
7.9 KB · Views: 413
Why not do the integral in terms of dy?

Well the professor asked us for bring both methods in terms of y as function of theta & theta as function of y;

I already did in terms of dy. making the substitution of cosine(θ) = x/√x^2+y^2 and then integrating.

but now i need the other method. y in terms of "θ"

OK - so complete: $$y=x\tan\theta\\ \implies \frac{dy}{d\theta}=\cdots$$ ... then multiply both sides by ##d\theta##.

Then substitute these values for y and dy into:
$$E_x=\frac{\lambda}{4\pi\epsilon_0}\int \frac{\cos\theta\; dy}{x^2+y^2}$$ ... with appropriate limits.

note: I wrote all that out so you'd get an example of LaTeX, since I noticed you struggling with typing out the math there.
To figure out how I did it - just hit the "quote" button attached to this post.
The equations all appear between dollar or hash marks.

I understand your confusion and frustration. The key here is to remember that when dealing with a charged line, the electric field is constant along the line and only varies with distance from the line. This means that the electric field can be expressed as a function of distance, rather than specific values of x and y.

To express y as a function of θ, we can use the trigonometric relationship between y and x, which is y = xtan(θ). This means that as θ changes, the value of y will also change. However, we cannot simply take the derivative of this equation to get dy/dθ, as y is a function of both x and θ.

To find dy/dθ, we need to use implicit differentiation. This involves taking the derivative of both sides of the equation with respect to θ, and then solving for dy/dθ. This will give us an expression for dy/dθ in terms of x and θ, which we can then use in our integral.

It's also important to note that in this case, we are integrating with respect to y, not θ. This means that we need to use the chain rule to express dy in terms of θ, and then substitute this into our integral.

Overall, it may seem complicated, but with a clear understanding of the relationships between distance, angle, and electric field, and using techniques like implicit differentiation and the chain rule, we can express y as a function of θ and successfully integrate to find the electric field due to a charged line.

## 1. What is an electric field?

An electric field is a physical quantity that describes the force experienced by a charged particle in the presence of other charged particles. It is a vector quantity, meaning it has both magnitude and direction.

## 2. What is a charged line?

A charged line is a theoretical line that has a certain amount of electric charge distributed along its length. It is often used as a simplification in mathematical models to describe the behavior of electric fields.

## 3. How is the electric field due to a charged line calculated?

The electric field at a point due to a charged line can be calculated using the equation E = kλ/ρ, where k is the Coulomb constant, λ is the charge per unit length of the line, and ρ is the distance from the point to the line. This equation assumes that the line is infinitely long and that the point is located on the perpendicular bisector of the line.

## 4. How does the electric field change with distance from the charged line?

The electric field due to a charged line follows an inverse relationship with distance. This means that as the distance from the line increases, the electric field decreases. The rate of decrease depends on the charge per unit length of the line and the distance from the line.

## 5. How does the electric field vary with angle from the charged line?

The electric field due to a charged line is strongest when the point is located on the perpendicular bisector of the line. As the angle from the line increases, the electric field decreases. This relationship follows a cosine function, with the electric field being zero when the point is located on the line (at an angle of 0 or 180 degrees) and reaching its maximum value at an angle of 90 degrees.

• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
64
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
404
• Introductory Physics Homework Help
Replies
23
Views
343
• Introductory Physics Homework Help
Replies
5
Views
732
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
791
• Introductory Physics Homework Help
Replies
1
Views
971