Electric Field due to Continuous Line Charge

In summary, the electric field of a point above a distance Z from a line is a cumulative Electric field caused by every charge in the line. To find the electric field at a point, you integrate the field across the line.
  • #1
darkpsi
23
0

Homework Statement


Find the electric field a distance z above the midpoint of a straight line segment of length
2L, which carries a uniform line charge λ.


Homework Equations


1) my textbook says :
E(r) = 1/4πεoV λ(r')/r2 r dl'


2) and this also works? :
E(r) = 1/4πεoV λ(r')/r3 r dl'

both where r is the unit vector from the charge to the point we are caluclating the field at


The Attempt at a Solution


I can integrate once I get the setup down just fine. I just wanted to know when and why this two are the same. For example, in my book it says for the setup for this problem :
E(r) = 2 * 1/4πεoV(λdx/r2)cosθz

For one thing, how did that cos appear and since it equals z/r couldn't 2) have just been used from the start? And if 2) can be used could I apply it to surface and volume integrals?
 
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  • #2
General formula for E is:

dE = 1/4πεo (dq/r^2)[tex]\hat{r}[/tex]

However, r = r.[tex]\hat{r}[/tex] (r_hat is unit vector)
So you can write"

dE = 1/4πεo (dq/r^3).r

--
Electric field of a point above distance Z from a line is a cumulative Electric field caused by every charge in straight line.

Lets say an angle formed by E caused by one charge and a vertical axis is theta. From there, you can get:
E_x = E * sin(theta)
E_y = E * cos(theta)

And because E in x-axis will automatically blow up, you just need to find E in y-axis.

Hope this helps
 
  • #3
Oh now I see. I wasn't even thinking that because r_hat wasn't pointing vertically that the z component wasn't all of the electric field. I was thinking the x-fields canceled so the field in the z direction was exactly equal to the electric field, but its actually only cos(theta) of it. Thanks a bunch
 

Related to Electric Field due to Continuous Line Charge

1. What is an electric field due to continuous line charge?

An electric field due to continuous line charge is a type of electric field that is produced by a line of charge with a consistent charge density. This means that the charge is spread out along an infinitely long line, rather than being concentrated at a single point.

2. How is the electric field strength calculated for a continuous line charge?

The electric field strength (E) for a continuous line charge is calculated using the equation E = (λ/2πε0) * ln(r/r0), where λ is the charge density, ε0 is the permittivity of free space, and r and r0 are the distances to the point of interest and the reference point, respectively.

3. What is the direction of the electric field due to a continuous line charge?

The direction of the electric field due to a continuous line charge is perpendicular to the line of charge and points away from the line for a positive charge and towards the line for a negative charge.

4. How does the electric field due to continuous line charge change with distance?

The electric field due to continuous line charge follows an inverse relationship with distance. This means that as the distance from the line of charge increases, the electric field strength decreases.

5. What are some real-life applications of the electric field due to continuous line charge?

The electric field due to continuous line charge has many real-life applications, including in the design of transmission lines for electricity, the behavior of charged particles in particle accelerators, and the development of capacitors and other electronic components. It is also important in understanding the behavior of lightning and atmospheric electricity.

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