Electric field in two dimensions

In summary, the problem is to find an expression for the electric field at point P, where the distance from P to the positive charged rod is Y. The solution involves breaking the charged rod into small differentials and integrating using the formula Ey = (integrate)-L to +L Q/(4pi Epsilon(o) d^2)dq. The X components can be calculated by using the formula dEx=dEcos(theta) and dEy=dEsin(theta), and substituting the charge density lambda*dx for dq. The radius, r, can be found using the formula sqrt(L^2+d^2), where d is the vertical distance from the line to the point. With some trigonometric changes, the integral for
  • #1
stunner5000pt
1,461
2
I have this problem which I am stuck on: Find an expression for the electric field at point P as in this figure
P
|
|
|
|
|
|
+++++++++++++++++++++++++++++++++++++++++++++++++++++++
<----------------------L-------------------------------->
Where the distance from P to the positive charged rod is Y.

All i know is that you need to break the positive charged rod into little differentials such as dL and integrate something like this -
Ey = (integrate)-L to +L Q/(4pi Epsilon(o) d^2)dq
However i am not sure about the X components and how taht is calculated, please help.
 
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  • #2
Why don't u apply Gauss Th
 
  • #3
Well this one can be tricky. Now with your picture there and then your claim I am confused, is the point located perpendicular to the end of the line of charge?
If that is the case then integrating along the length from 0 to L might be easier...
anyway
if you look at the picture, as you move from the left to the right, and angle is formed with the line extending from P along L. as you move down L this angle changes.
As the angle becomes larger the radius changes as well.
dEx=dEcos(theta)
dEy=dEsin(theta)
Now dE=Ko*dq/(r^2)
substituting the charge density lambda*dx for dq we get
dE=Ko*Lambda*dx/(r^2)
if you replace r with the varibles we do know you get
r=sqrt(L^2+d^2)
where d is the vertical distance from the line to the point.
now making the necessary trigonometric changes you can find the the integral for each component.

Hope that helps
 

Related to Electric field in two dimensions

1. What is an electric field in two dimensions?

An electric field in two dimensions refers to the force exerted on a charged particle by other charged particles in a two-dimensional space. It is a vector field that describes the strength and direction of the force at each point in the space.

2. How is the electric field calculated in two dimensions?

The electric field in two dimensions can be calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. What are the units of an electric field in two dimensions?

The units of an electric field in two dimensions are Newtons per Coulomb (N/C) or Volts per meter (V/m). These units represent the strength of the electric field at a particular point.

4. How does the direction of an electric field change in two dimensions?

The direction of an electric field in two dimensions depends on the location and charge of the particles in the space. The electric field lines always point away from positive charges and towards negative charges.

5. How does the presence of conductors affect the electric field in two dimensions?

Conductors, such as metals, have free electrons that can move in response to an electric field. This causes the electric field to be zero inside a conductor, as the charges redistribute themselves to cancel out the field. However, the electric field outside the conductor is not affected.

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