# Magnitude of force on point from wire

1. Mar 3, 2015

### Ozmahn

1. The problem statement, all variables and given/known data
A straight, nonconducting plastic wire 9.50cm long carries a charge density of 100 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.

Find the magnitude and direction of the electric field this wire produces at a point 6.00cm directly above its midpoint.

If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00cm directly above its center.

$\lambda=100 {\frac{nC}{m}}$

2. Relevant equations

Area of cylinder
$A=\pi r^2*h$
$\oint \vec E \cdot d \vec A = {\frac{Q_{enclosed}}{\epsilon_0}}$

3. The attempt at a solution
Create a cylinder around wire, with end caps at either end of the wire.

$\oint \vec E \cdot d \vec A = {\frac{Q_{enclosed}}{\epsilon_0}}$

since E is perpendicular to dA

E is constant at the radius of P (.06m), so you can factor it out of the integral.

$\ E \oint \cdot dA$

Integrating dA gives you A. Area of a cylinder is given by

$A= \pi r^2*h$

$E* \pi r^2*h={\frac{Q_{enclosed}}{\epsilon_0}}$

Since Q is the same as the charge density ($lambda$) times length,

$Q=\lambda*0.095$

The height of the cylinder is equivalent to the length of the wire, so

$h=l=0.095$

Sub that into Gauss' law, and I have

$E= {\frac{\lambda*l}{\pi*r^2*\epsilon_0 (l)}}$

The lengths (0.095) cancel, and I separate the k constant from the rest

$E= {\frac{\lambda}{r^2}} * {\frac{1}{\pi \epsilon_0}}$

${\frac{1}{\pi \epsilon_0}}=4*{\frac{1}{4\pi \epsilon_0}} = 4k = 4*8.99*10^9$

Plugging this in, my equation looks like

${\frac{100*10^{-9}*(4*8.99*10^9)}{0.06^2}} = 998888.8889$

That's not the right answer, but I'm not sure where I messed up. Any help is appreciated, thanks.

2. Mar 3, 2015

### TSny

Hi, Ozmahn.

You treated E as constant over the cylindrical Gaussian surface (you pulled it out of the integral). If the wire were infinitely long, that would be correct. Can you see that for a finite length of wire, E has different magnitudes and directions for different patches of area dA on the surface?

If you convince yourself that there is not enough symmetry in this problem to use Gauss' law, how else might you approach the problem?

3. Mar 3, 2015

### Ozmahn

I think I can see what you're saying. Basically, the electric force isn't moving in the radial direction relative to the wire?

4. Mar 3, 2015

### TSny

By "moving" do you mean "pointing"? (The electric field is not something that moves in this problem.) But I think you have the right idea. At most points around the wire, The electric field does not point in a direction perpendicular to the wire.

5. Mar 3, 2015

### Ozmahn

Okay got it. So then no Gaussian surfaces since there is no symmetry in this case. So the formula
$E = {\frac{kQ}{r^2}$
Could be used...that would give me the electric field at the point. And Q would be the charge density times the length of the wire...I think.

6. Mar 3, 2015

### TSny

No, you cannot treat the entire wire as a single point charge. You will need to break up the wire into infinitesimal sections and integrate. Hopefully you have done similar examples in class.

7. Mar 4, 2015

### Ozmahn

Oh we did do an example of that in class. Okay so starting over...

The wire will be broken up into infinitely small segments called dL. dQ will be $\lambda dy$, and r will just be the distance from the center of the wire to point P, which is 0.06 away. I didn't understand this next part very much, but here is what I think I need:

$-E_y=E_y$ cancelled due to symmetry
$E_x=\int dE_x$
$dE= {\frac{1}{4 \pi \epsilon_0}} * {\frac{\lambda dy}{r^2}}$
$dE_x= {\frac{1}{4 \pi \epsilon_0}} * {\frac{\lambda dy}{r^2}} * cos(\theta)$

I'm following this so far, but in this last step he added $cos(\theta)$ to account for the angle. The angle created by the point is 90 degrees, so wouldn't that make cos90=0 and everything would be 0?

8. Mar 4, 2015

### TSny

A small segment $dy$ of the wire that is located at some value $y$ on the y-axis produces an electric field $\vec{dE}$ at the $P$ located 6 cm from the origin on the x-axis. $\vec{dE}$ makes an angle $\theta$ with respect to the x-axis that depends on the location of $dy$.

9. Mar 4, 2015

### Ozmahn

I'm having some real issues with this one..
I started over again, so here is what I've got so far:

I know the y components cancel, so I'm left looking for the sum of all the x components.
I'm switching from dy to dx to make things easier for myself, since the distance along the wire can also be referred to as x.

$dE_x=dEcos(\theta)$
$dE={\frac{kdQ}{r^2}} cos{\theta}$
$\lambda = {\frac{dQ}{dx}} = dQ = \lambda dx$
$dE={\frac{k \lambda dx}{r^2}} cos{\theta}$
$r= \sqrt{0.06^2+x^2}$
$\int dE = E$
$\int {\frac{k \lambda dx}{r^2}} cos{\theta}$
bounded from 0 to 0.095
Factor out constants, plug everything in
$k \lambda \int {\frac{1}{0.06^2+x^2}} cos{\theta} dx$
Find what $cos(\theta)$ is in terms of dx
$cos( \theta) = {\frac{A}{H}} = {\frac{0.06}{\sqrt{0.06^2+x^2}}}$
$k \lambda \int {\frac{1}{0.06^2+x^2}} {\frac{0.06}{\sqrt{0.06^2+x^2}}} dx$
Combine inside integral
$k \lambda \int {\frac{0.06}{(0.06^2+x^2)^{3/2}}} dx$
Factor out 0.06 from denominator and cancel
$k \lambda \int {\frac{1}{(1+{\frac{x^2}{0.06}})^{3/2}}} dx$
still bounded from 0 to 0.095
After integrating, I got 0.0885719. But the answer I'm getting for E is 79.6. I doublechecked for algebraic errors, but I don't see any. Do I have the right idea or am I still totally off the ball with this?

10. Mar 4, 2015

### TSny

Your limits of integration are not correct. Also, check the last step where you factored out .06. Otherwise, it looks good.

11. Mar 4, 2015

### Ozmahn

I put the .06 back in there and integrated over the bounds -.0495 and .0495, but I'm pretty sure those bounds are wrong too because I get 4671.25N/C. The bounds should be the length of the wire, shouldn't they?

12. Mar 4, 2015

### TSny

Yes, the bounds should cover the entire wire. Keep in mind where x = 0 is located on the wire.

Why 0.0495? What is half of 0.095?

13. Mar 4, 2015

### Ozmahn

I must have written that on my paper by accident, it should actually be 0.0475. I used the number because that's how my professor did the problem, but if I'm understanding what you're saying, the position of the object is x=0. And since it's centered directly above the center of the wire, 0.095 divided into two equal sections would have half the values negative and half the values positive. So the bounds should be -0.0475, 0.0475. And that solved it! Thanks a lot for your help!! I'm definitely understanding these problems a lot more.