A Semi- Infinite Conducting Rod

[dunna]

Homework Statement

25. A semi-infinite nonconducting rod (i.e. infinite in one
direction only) has uniform positive linear charge density
lambda. Show that the electric field at point P makes an
angle of 45 degrees with the rod and that this result is
independent of the distance R. (HINT: Separately find
the parallel and perpendicular (to the rod) components
of the electric field at P, and then compare those
components.)

rod
starts
here
_ +++++++++++++++++++++++++ =====> very long
| |
|
R |
|
| |
- P <== this point is a distance R from the end
of the rod

Homework Equations

Coulomb's Law
E= kq/r^2

Q= (lambda)*x and thus dQ=(lambda)*dx (assuming the semi-infinite rod beings at 0 and continues on the x-axis)

The Attempt at a Solution

I drew a diagram and using the problems suggestions I solved for the perpendicular component first which I called dE_y (assuming it ran with the y-axis)
dE= (kdQ)/R^2

Then solving for the "parallel" component, I understand that if the rod truly continues to infinity, the distance between the point and the rod will become negligible and thus the influence by the rod on the point will truly be parallel with the rod itself.

dE_x = dEcos(theta) = (k*lambda*dx*cos(theta))/r^2
= (k*lambda*dx*y_0)/(y_0^2+x^2)^(3/2)

and from here it all becomes convoluted.

Thank you for your time

 

[anathema]

and help.

Thank you for your question. I can see that you have made a good attempt at solving this problem, but there are a few errors in your solution. Let me guide you through the correct approach.

Firstly, let's consider the electric field at point P due to a small element of charge dQ located at a distance x from point P. By Coulomb's law, the magnitude of the electric field at P due to this element is given by:

dE = k*dQ/R^2

However, we need to consider the direction of this electric field. Since the element of charge is located along the x-axis, the electric field will be directed along the line joining the element and point P. This line makes an angle theta with the x-axis, as shown in the diagram. Therefore, the x- and y-components of the electric field at P are given by:

dE_x = dE*cos(theta) = (k*dQ*x)/(R^2*(x^2+R^2)^0.5)

dE_y = dE*sin(theta) = (k*dQ*R)/(R^2*(x^2+R^2)^0.5)

Note that the x-component of the electric field is parallel to the rod, while the y-component is perpendicular to the rod.

Now, we need to integrate these components over the entire length of the rod to find the total electric field at point P. Since the rod is infinitely long, we can consider only the positive x-axis. The limits of integration are from x=0 to x=∞.

E_x = ∫ dE_x = ∫ (k*lambda*dx*x)/(R^2*(x^2+R^2)^0.5) = (k*lambda/R)*∫ (dx)/((x^2+R^2)^0.5)

= (k*lambda/R)*[sinh^-1(x/R)]0∞ = (k*lambda/R)*π/2 = (k*lambda/2R)

Similarly, E_y = ∫ dE_y = ∫ (k*lambda*dx*R)/(R^2*(x^2+R^2)^0.5) = (k*lambda/R)*∫ (dx)/((x^2+R^2)^0.5)

= (k*lambda/R)*[cosh^-1