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

[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;

 

[Simon Bridge]

Why not do the integral in terms of dy?

 

[Molderish]

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"

 

[Simon Bridge]

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.

 

[HalfLostAndSearching]

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.