# Two problems gettin on my nerve

4. A thin rod of length ℓ and uniform charge per unit length λ lies along the x axis, as shown in the Figure. (a) Show that the electric field at P, a distance y from the rod along its perpendicular bisector, has no x component and is given by E = 2ke λ sin θ0/y. (b) What If? Using your result to part (a), show that the field of a rod of infinite length is E = 2ke λ /y. (Suggestion: First calculate the field at P due to an element of length dx, which has a charge λ dx. Then change variables from x to θ, using the relationships x = y tan θ and dx = y sec2 θ dθ , and integrate over θ.)

for this problem, im not gettin how u arrive at the E field. (ie part a)

the other problem:

7. A line of charge with uniform density 35.0 nC/m lies along the line y = 15.0 cm, between the points with coordinates x = 0 and x = 40.0 cm. Find the electric field it creates at the origin.

i get the value but not the irection of the E field.

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Now for your other problem, you can perform a vector summation pretty easily. just sum it up like Ex=int((Kdq/distance^2)sin@) where @ is the angle between the line x=0 and the line from the origin to the charge dq. find a way to sum this from one side of the bar to the other. Then get your y-component and then you should have no problem getting your direction.

I proved that E has no x component but i cant find the y component . can some1 show me a procedure (specially help withg the integration part)

as for ur suggestion for the other problem. i dont think im familiar with the kind of math ur using... it loox weird

HallsofIvy
Homework Helper
Once again, what have you DONE to set up the integral. Show us what you have tried.

ooo ok (i will denote lambda by @

dq = @ dx

dE = k @ dx / r^2 = k @ dx /(x^2+y^2)

E = int (k @ dx / (x^2 + y^2)) from -.5L to .5 L

i cant integrate it and i cant find the y component
too

pls also help me out for the other problem. i get the general notion of it already, but its the math where i go wrong. show me the INTEGRATIOn

cristo
Staff Emeritus
Have you tried the substitution given in the hint in the question?

Hello all! I'm trying to work on this problem as well but I'm a bit lost.
So far, I know that the Ex component will cancel, so only Ey will remain.
Then I know that dq = lamda dx.
I'm defining theta as the angle adjacent to y, so cos(ᶿ)= y / [√(L/2)^2 + (y)^2], where [√(L/2)^2 + y^2] = r.
and I know that dE = (kdq) / r^2

Substituting that into the equation for dE gets me:
dE = k (lamda dx) / (r)^2 * (y / r)

From here we can take an integral to get E, so:
E = k(lamda)y ∫ dx / ((L/2)^2 + (y)^2)^1.5
E = k(lamda)y (L/2) / y √((L/2)^2 + y^2)
E = k(lamda)L / 2y√((L/2)^2 + y^2), integrating from L to 0?

I'm actually not sure if integrating from L to 0 is correct... when I tried it I got
(k(lamda)L) / (2√((L^2/4) + L^2) - k(lamda)
which does not resemble the final answer, E = 2k*lamda / y

Any guidance at all would be most appreciated. Thanks!