is my solution correct ..

Nov 2008
1,401
2
here is a question:
http://i25.tinypic.com/8xlcfl.gif
there is endless wire uniformly charged \(\displaystyle \lambda=2*10^{-4}C/m\)
and close to it located a wire with \(\displaystyle l=0.12_m\)
which has 30 degree angle between him and the endless wire,and it charged in uniformed way so that the total charge on it is \(\displaystyle q=3*10^{-9}\)
the distance of the finite wire and the infinite wire is 0.08 meters

A.find the total force acted on the finite wire.
B.find it when \(\displaystyle \alpha=0 \)degrees and when \(\displaystyle \alpha=90 \)degrees.

gaus law for infinite wire i presume its length is l and distance r.
\(\displaystyle E2\pi r h=\frac{q}{\epsilon_0}\\\)
\(\displaystyle E=\frac{q}{\epsilon_0 2\pi r h}=\frac{\lambda h}{\epsilon_0 2\pi r h}=\frac{\lambda }{\epsilon_0 2\pi r }\\\)
\(\displaystyle \frac{q_2}{l_2}=\lambda_2\\\)
\(\displaystyle l_0=l_2\cos \alpha\\\)
\(\displaystyle \int_{0}^{l_0}Edl=\int_{0}^{l_0}\frac{\lambda }{\epsilon_0 2\pi r }dl\)

i sum along the finite wire
but the ditance from the infinite changes too.
so it should be a double integral
but its not a 2d body

??
 
Last edited:
Jul 2009
68
20
you are right, if its a 1 dimensional body you will need only 1 integral

but you also have to find out the relation of r with l.


hint:
if you draw a xy coordinate axis (parallel and perpendicular to the infinite wire), you can do it

then
\(\displaystyle dl=\sqrt{dx^2+dy^2}\)
\(\displaystyle dl=\sqrt{(\frac{dx}{dx})^2+(\frac{dy}{dx})^2}dx\)
\(\displaystyle dl=\sqrt{1+(\frac{dy}{dx})^2}dx\)
and you got a 1d integral
 
Nov 2008
1,401
2
i cant use expression using dx and dy
because thus i wil have a double integral

i need to link l with r
r=0.08+(temporary hipotenuse)*cos alpha

i dont know how to express this temporary hipotenuse
 
Jul 2009
68
20
i cant use expression using dx and dy
because thus i wil have a double integral
no.
its not dx and dy. its dx and \(\displaystyle \frac{dy}{dx}\).


Imagine these axis. y-axis parallel the infinite wire. x-axis perpendicular.
that way r will simply be x.

now you need to integrate dl, as you wrote in the end of your 1st post. but you need to write dl into a function of dx.




notice the 90º angle triangle, you use





now you just need to find the wire equation
y=mx+b
calculate \(\displaystyle \frac{dy}{dx}\) and you get dl in terms of dx.

Then substitute dl in your equation.
Actually your equation isnt entirely correct (Ithink)

F=E q
so you need to use
\(\displaystyle dF=E \times dq\)
\(\displaystyle dF=E \times \lambda_2 \times dl\)
 
Nov 2008
1,401
2
my expestion is correct because its F=dq*E

and we substitute dl by an expression of dx
but i cant express dy only with dx
\(\displaystyle
\int_{0}^{l_0}Edl=\int_{0}^{xpretion}\frac{\lambda }{\epsilon_0 2\pi r }\sqrt{1+(\frac{dy}{dx})^2}dx

\)
 
Jul 2009
68
20
y=mx+b

if you derive that in relation to x

\(\displaystyle \frac{dy}{dx}=m\)

you look to the line and you see its 0.12m long and makes 30º with vertical, so
the x component of the line, \(\displaystyle \Delta x\), is
\(\displaystyle \Delta x=0.12 \times sin(30)=0.06\) m

the y component of the line, \(\displaystyle \Delta y\), is
\(\displaystyle \Delta y=0.12 \times cos(30)=0.06 \sqrt 3 \) m

\(\displaystyle m=\frac{\Delta y}{\Delta x}=\sqrt 3\)

\(\displaystyle \frac{dy}{dx}=\sqrt 3\)



my expestion is correct because its F=dq*E

and we substitute dl by an expression of dx
but i cant express dy only with dx
I don't understand.
if you say F=dq*E, why do you use in the integral E dl.
If you are calculating the Force you need to use E dq.

Note that nowhere in your expression the electric charge of the finite line is included. And the force clearly depends on that charge.
 
Nov 2008
1,401
2
\(\displaystyle
E=\frac{\lambda }{\epsilon_0 2\pi r }\\\)
\(\displaystyle
\int_{0}^{0.12}Edq=\int_{0}^{0.12}\frac{\lambda^2\sqrt{1+3}dx }{\epsilon_0 2\pi r }=\int_{0.08}^{0.08+0.06 \sqrt 3}\frac{\lambda^22dx }{\epsilon_0 2\pi r }=\int_{0.08}^{0.08+0.06 \sqrt 3}\frac{\lambda^22dx }{\epsilon_0 2\pi x }\)

correct?