Homework Help: Find the electric field produced by two rods

1. Jan 21, 2012

kasparov5

1. The problem statement, all variables and given/known data
Two thin rods of length L lie on the x axis. One of them has a charge of
-Q uniformly distributed along its length and is positioned between x = -L
and x = 0. The other one has a charge of +Q uniformly distributed along
its length and is positioned between x = 0 and x = L. Find the electric field
produced by the two rods (magnitude and direction) at a point O located on
the y axis at y = h.

2. Relevant equations

I would imagine the relevant equation is E=kq/r^2.

I would kind of like to solve this without calculus, since i don't understand that.

3. The attempt at a solution

I think the vertical components cancel out. After that, I really have no idea what to do. Maybe you plug in the distance into the above formula. I am really confused with this problem.

2. Jan 21, 2012

Staff: Mentor

Good.
The formula you quoted is for a point charge.

I see no way to avoid using a bit of calculus here.

3. Jan 21, 2012

kencoumerilh

I won't address your question, but I do recommend you buy a simple book on calculus for nighttime reading and enjoyment. When I was young, I had a series of books on mathematics, one of which was "calculus for the practical man". I own several calculus text books now, but this little book taught me calculus in 7th grade and I understood what I was doing, not just memorizing. Calculus is easy if you understand why you are using it. Learn differentials, integrals and differential equations, which isn't in that little book, but very useful in life.

Good luck with your question ...

Ken

4. Jan 21, 2012

kasparov5

I think I did it without calculus. After realizing that the vertical components cancel out. I then realized that the net electric field produced by both rods will be simply 2 times the horizontal component of the electric field produced by one of the two charges. After some plugging and chugging and trigonometry, I got the following answer:

E_net on point O = (2KQL)/(((L^2)+(h^2))^3/2)

Is this correct? Is my logic sound? If not, what could I be missing.

5. Jan 21, 2012

Staff: Mentor

Good. But realize that each charge is a line of charge, not a point charge. (You'll have to integrate.)
How did you arrive at this answer?

6. Jan 21, 2012

kasparov5

I arrived at that answer by converting cos theta to variables L and h and I multiplied that by twice the electric field caused by the charge at the very end of the bar. Clearly, I forgot the fact that it is indeed a LINE of charge.

Can you show me how to begin the integration for this problem? I really have no idea.

7. Jan 21, 2012

Staff: Mentor

Break up the line of charge into small elements of length dx. The charge density would be λ = Q/L. Then express the horizontal component of the field due to an element of charge at position x. Set that up, then integrate to get the total field from the entire line of charge. Give it a shot.

8. Jan 21, 2012

kasparov5

I did that. I removed the constants 2k(lambda) (2 coming from multiplying by 2 since there are two bars) and i have an integral from the length L to 0 of (x)/(x^2 +h^2)^3/2 dx.......but i do not know how to integrate this. There is an h inside there that technically is also a constant, but I cannot get it out of the integral like 2k(lambda). How do I integrate this last section?

Thanks!

9. Jan 21, 2012

Staff: Mentor

Just play around with it a bit. Try a substitution, such as y^2 = x^2 + h^2.

10. Jan 21, 2012

kasparov5

I think I got the answer as being E_net = 2k(Q/L)*[(1/h)-(1/sqrt(L^2+h^2))] where Q/L is of course lambda. I got this by substituting like you said and then plugging in L and 0 for x to find the definite integral. The electric field is in the negative x direction.

Please tell me if this is the correct answer. I worked quite hard on it. Is there anything I did wrong?

11. Jan 21, 2012

Staff: Mentor

Looks good to me. Nice job!

12. Jan 21, 2012

kasparov5

oh no....i think it is wrong because what if (1/h)-(1/sqrt(L^2+h^2)) turns out negative????

I might get a positive or negative answer.

What does this mean? Is it wrong?

13. Jan 21, 2012

Staff: Mentor

Do you think that's possible? Compare h to sqrt(L^2+h^2).

14. Jan 21, 2012