# Electric Field Intensity of a line on bisecting plane

Tags:
1. Aug 4, 2017

### elements

1. The problem statement, all variables and given/known data
A very thin, finite, and uniformly charged line of length 10 m carries a charge of 10 µC/m. Calculate the electric field intensity in a plane bisecting the line at ρ = 5 m.
2. Relevant equations

3. The attempt at a solution

Not sure why i'm not getiting this but i've been at this for 8 hours now and I still cannot figure out how to solve it. So far i've attempted to use various types of surfaces to see if there was something I could do to calculate a point charge on the plane in order to integrate it across the surface.ρ refers to the radial component in cylindrical coordinates. I've attached a picture of one of my attempts at drawing the situation. My other attempt is the method of vector addition, where the x components cancel out.

My attempts thus far:

$$\oint \vec E \cdot \vec {dA}=Q_{encl}/ε_0$$
$$\vec E \oint_S {\vec{dA}}= Q_{encl}/ε_0$$
$$(2πρ_lL)E = ρ_l/ε_0$$
$$E=ρ_l/2πρε_0$$
$$1*10^-6/2π(5)*9x10^9=7.2*10^5 C$$
Other method plot

The other approach I took was the to deal with it in vector form, although i'm still not exactly sure what i'm being given with the ρ=5m.
Known Variables:
$$\vec ρ = (3,4,5)$$
$$r=\sqrt{x^2+y^2+z^2}=7.1m$$
$$q = 10*10^{-4} C/m$$
$$\vec p = q \vec d$$
$$\vec p =(10*10^{-6} C/m \hat k)(10 \hat k)$$
$$\vec E = -\vec{\nabla} \cdot \vec V$$
$$\vec E = (3(\vec p \cdot \vec r)(\vec r) - r^2 \vec p )/ 4πε_0(r^5)$$
$$\vec E = 3((10^-4 \hat k)\cdot (5 \hat k)(3 \hat i + 4 \hat j +5 \hat k)-7.1^2(4*10^-3)/(4πε_0)(7.1)^5$$
$$\vec E = 10^3 \cdot (2.2 \hat i + 3.0 \hat j + 1.2 \hat k) V/m$$

I am unsure what else I can do I can't figure out exactly how to get the field intensity of the plane. - convert to cylindrical somehow for the divergence equation?

#### Attached Files:

• ###### Untitled2.png
File size:
11.4 KB
Views:
19
Last edited: Aug 4, 2017
2. Aug 5, 2017

### ehild

When using Gauss Law you have to integrate to a closed surface, that is, the bottom and top plates of the cylinder have to be included. This is not a road of infinite length! The point where you have to calculate the electric field is at distance half the length of the rod. And you can not assume that the electric field is perpendicular to the surface of the cylinder, or it is of equal magnitude along the surface.
And what do you mean on the numerical value of 7.2*10^5 C? C is the unit of charge, and you want electric field.
What is your $$\vec ρ = (3,4,5)$$?
There is no dipole, positive charge is distributed uniformly along the rod.
Take a small length dl considering it a point charge of (Q/L)dl and determine its contribution to the electric field on the bisecting line at ρ=5 m from the rod. Integrate for the full length.

3. Aug 5, 2017

### elements

$$\vec ρ = (3,4,5)$$ that would be my radius vector that I was using for the r, and what about the plane? Are we not looking for the field across the entire plane?

4. Aug 5, 2017

### ehild

We are looking for the electric field at all points of the plane, that are at 5 m distance from the rod. It is a circle, not a fixed vector. And the charge from all along the rod contributes to it.