# Electric Field and Infinite Plane

1. Jul 11, 2006

### verd

Hi,

I've worked out a problem and don't understand why it's wrong... I think I might be going about this incorrectly.

Here it is:

http://www.synthdriven.com/images/deletable/01.jpg [Broken]

I started this by attempting to solve for E. In the beginning of the problem, I'm told that this is an infinite plate, "a uniformly charged vertical sheet of infinite extent", and have determined the following formula:

$$\vec{E}=\frac{\sigma}{2\epsilon_{o}}$$
(This is what was derived in class for an infinite surface, I can go over that if anyone would like... sigma=Q/A)

I'm taking the "areal charge density" to be sigma.

$$\vec{E}=\frac{\sigma}{2\epsilon_{o}}=\frac{0.12\times10^{-6}}{2(8.85\times10^{-12})}=6779.66N/C$$

Using the following formula, I determined the force, (where q is my given point charge):

$$F=E\timesq=(6779.66)(0.11\times10^{-6})=0.000746N$$

http://www.synthdriven.com/images/deletable/02.jpg [Broken]

According to my little force diagram, this F is equivalent to my x-component of the force labelled F. And the y-component of this force is zero...

The next force is the weight, mg. I wasn't sure whether or not to do this in grams or kilograms, so I kept it at grams (my professor converted from kg to g in a previous example in class... even though some problems are typically answered in kg)

$$W=mg=(1g)(9.8m/s^2)=9.8N$$

This force will be only in the -y direction, obviously.

And the tension, T is as follows:

$$T_{x}=T\sin{\theta}$$
$$T_{y}=T\cos{\theta}$$

$$\sum{F_{NETx}}=F_{x}+W_{x}-T_{x}=0.00076N+0N-T\sin{\theta}=0$$
$$T\sin{\theta}=0.00076N$$
$$T=\frac{0.00076N}{\sin{\theta}}$$

$$\sum{F_{NETy}}=F_{y}-W_{y}+T_{y}=0-9.8N+T\cos{\theta}=0$$
$$T\cos{\theta}=9.8N$$
$$T=\frac{9.8N}{\cos{\theta}}$$

$$\frac{0.00076N}{\sin{\theta}}=\frac{9.8N}{\cos{\theta}}$$
$$\frac{\sin{\theta}}{\cos{\theta}}=\frac{0.00076}{9.8}$$
$$\tan{\theta}=\frac{0.00076}{9.8}$$
$$\arctan{\frac{0.00076}{9.8}}=\theta$$
$$=0.004443$$ degrees

What am I doing wrong? This seemed like the right way to go about it...

Last edited by a moderator: May 2, 2017
2. Jul 11, 2006

### neutrino

Your units have to be consistent. Since you are using S.I., the unit of mass is Kg, and therefore the mass of the object is 10-3Kg.

Remember this, the S.I units are also known as the MKS system (Metre, Kilogram, Second). The cgs stands for (Centimetre, Gram, Second).

http://en.wikipedia.org/wiki/Mks

3. Jul 11, 2006

### verd

Thanks.

That'd make my answer 4.3447 degrees... And yet my answer is still incorrect... Any ideas?

4. Jul 11, 2006

### neutrino

I cannot spot anything wrong in your calculation. I wonder why the length of the string has been provided (probably just to make you go off-track).

5. Jul 11, 2006

### verd

Yeah... That's what I was thinking. But who knows. I could actually have to use it somehow.

Seems like a way to make you waste your time realizing that you don't have enough information to find the angle trigometrically w/out physics.

6. Jul 11, 2006

### arunbg

7. Jul 11, 2006

### verd

It's an online homework service. I have a couple of chances at an answer... So I don't know what the answer is.

8. Jul 11, 2006

### arunbg

Well, was any option close to the answer at least ?

9. Jul 11, 2006

### verd

Nope. I have no idea what it is. This method seems correct.

10. Jul 11, 2006

### neutrino

May be (some of) the numbers you have are wrong. For example, that 0.00076 should be 0.000745/6. Just make sure that the numbers are correct to a few decimal places and submit an answer. Either that, or the online service has totally gone berserk. ;)