Electric Field Intensity of a line on bisecting plane

In summary, the conversation discussed the calculation of the electric field intensity at a plane bisecting a uniformly charged line. Various attempts and equations were presented, including Gauss's Law and vector calculations, but the correct approach was to consider the contribution of small lengths of the line to the electric field at the point of interest and integrate for the full length. The final answer should be a circle, not a fixed vector, and the charge from all along the rod contributes to the electric field intensity at the plane.
  • #1
elements
29
0

Homework Statement


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.

Homework Equations

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:
Untitled.png


$$\oint \vec E \cdot \vec {dA}=Q_{encl}/ε_0$$
$$\vec E \oint_S {\vec{dA}}= Q_{encl}/ε_0$$
$$(2πρ_lL)E = ρ_l/ε_0$$
$$E=ρ_l/2πρε_0$$
$$1*10^-6/2π(5)*9x10^9=7.2*10^5 C$$
Other method plot
Untitled2.png

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?
 

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  • #2
elements said:

Homework Statement


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.

Homework Equations

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:
View attachment 208326

$$\oint \vec E \cdot \vec {dA}=Q_{encl}/ε_0$$
$$\vec E \oint_S {\vec{dA}}= Q_{encl}/ε_0$$
$$(2πρ_lL)E = ρ_l/ε_0$$
$$E=ρ_l/2πρε_0$$
$$1*10^-6/2π(5)*9x10^9=7.2*10^5 C$$
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.
elements said:
Other method plot
View attachment 208333
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?
Calculate the electric field intensity in a plane bisecting the line at ρ = 5 m.
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
ehild said:
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.

$$\vec ρ = (3,4,5)$$ 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
elements said:
$$\vec ρ = (3,4,5)$$ 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?
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.
upload_2017-8-5_14-36-9.png
 

Related to Electric Field Intensity of a line on bisecting plane

1. What is the formula for calculating the electric field intensity of a line on a bisecting plane?

The formula for calculating the electric field intensity (E) of a line on a bisecting plane is E = (2kλ)/r, where k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), λ is the charge density of the line, and r is the distance from the line to the point where the electric field is being calculated.

2. How does the distance from the line affect the electric field intensity on a bisecting plane?

The electric field intensity is inversely proportional to the distance from the line. This means that as the distance increases, the electric field intensity decreases. This relationship is described by the formula E = (2kλ)/r, where r is the distance from the line.

3. What is the direction of the electric field intensity on a bisecting plane?

The direction of the electric field intensity is always perpendicular to the bisecting plane, pointing away from the line in both directions. This is because the electric field lines of a line charge are always perpendicular to the line itself.

4. How does the charge density affect the electric field intensity on a bisecting plane?

The electric field intensity is directly proportional to the charge density of the line. This means that as the charge density increases, the electric field intensity also increases. This relationship is described by the formula E = (2kλ)/r, where λ is the charge density of the line.

5. Can the electric field intensity on a bisecting plane be negative?

Yes, the electric field intensity on a bisecting plane can be negative. This indicates that the direction of the electric field is in the opposite direction as the conventionally defined positive direction. It is important to note that the magnitude of the electric field intensity is always positive, but the direction can be either positive or negative.

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