[symbolic] Linear uniform charge density (E field at a point)

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
syntroniks
Messages
3
Reaction score
0

Homework Statement


Center a rod of length L at (0,0) with the length going horizontally.

Take a point P at (0,y).

Find the electric field at P.

Homework Equations


[tex]\lambda[/tex]=Q/L
[tex]E= \int k*dQ/R^{2}[/tex]

The Attempt at a Solution


I am integrating from -L/2 to L/2
Since Q=lambda*L, I guess differentially dQ=lambda*dL.

Substituting that into the integral, it becomes:
[tex]k*\lambda \int dL/R^{2}[/tex]
from -L/2 to L/2 of course.

R is pretty messy so I'll just write what I came up with for [tex]R^{2}[/tex]:
[tex]R^{2}=(L^{2}/4)+y^{2}[/tex]

So... Doesn't this seem pretty reasonable? I just want to be double sure that this is OK.
 
Physics news on Phys.org
There may be a confusion with L, used in two different ways here.
I suggest you change dQ=lambda*dL to dQ=lambda*dx, where x is a distance along the x axis. This x runs from -L/2 to L/2.
I think you'll find that R² = x² + y².
Looks like one of those trig substitution integrals.
 
Thankfully for this particular problem I get to do the integration by software... kinda. Turns out to involve arctangent and a relative mess of symbols.

Thanks for the suggestion about variables, it is definitely more clear that way.
 
Looks like the substitution x = y*tan A really simplifies it!
And the A is a real angle in the problem.