What is the formula for the electric field above a plane?

In summary, the conversation is about solving a physics problem involving a uniformly charged wire of length L and finding the electric field at a point P, which is a perpendicular distance x from the midpoint of the wire. The problem involves using the equation E= (lambda/2*pi*epsilon_0) *(L/x*sqrt(L^2+4x^2)) and integrating cos theta with respect to dl, but there is a discrepancy in the results. The correct solution involves using the substitution cos theta = x/sqrt(x^2+(l/2)^2).
  • #1
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Homework Statement


This is problem 47(chapter 21) in the textbook - Physics for engineers and scientists (Giancoli)
Uniformly charged wire has length L, where point 0 is the mid point. Show that the field at P, perpendicular distance x from 0 is

E= (lambda/2*pi*epsilon_0) *(L/x*sqrt(L^2+4x^2)

Homework Equations

The Attempt at a Solution



I tried solving it, I got E = - (lambda/2pi*epsilon_0)*[1/sqrt(L^2+4x^2)-1/2x)
Is something wrong with my integration?

My attempt is correct until

E= lamda/r^2*4pi*epsilon_0 * integration (cos theta)dl

After this, I use cos theta = x/r (r is the hypotenuse = sqrt (L^2+4x^2))

I am trying to do this instead of taking r=x cos theta.
 
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  • #2
Looks like something wrong with the "(cos theta)dl."
I'll use A instead of theta. And z in place of your l, running from -L/2 to L/2.
I get some constants times integral of cos(A)/(x^2 + z^2)*dz
Since tan(A) = z/x, I can use z = x*tan(A) to simplify the integral.
And dz = x*sec^2(A) dA
After the dust settles on this change of variable, I get integral of cos(A)dA.

Not the difference from your integral: I have dA where you have dl
I end up with the given answer.
 
  • #3
If I use r=cos theta/x, then i get the correct result.
I am trying to get there without that substitution and just cos theta = x/sqrt(x^2+(l/2)^2)
huh. Doesn't work if I use the integral but does work if I take l/2 as y initially. The whole term becomes (x^2+y^2) and then apply limit to get answer.
 

1. What is an E field above a plane?

The E field above a plane refers to the electric field that exists in the space above a flat surface. It is a vector quantity that represents the strength and direction of the electric force experienced by a charged particle in that region.

2. How is the E field above a plane calculated?

The E field above a plane can be calculated using the formula: E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space. Alternatively, it can also be calculated by taking the derivative of the electric potential with respect to distance above the plane.

3. What is the direction of the E field above a plane?

The direction of the E field above a plane is perpendicular to the surface of the plane. This means that the electric field lines will point away from the plane if it is positively charged and towards the plane if it is negatively charged.

4. How does the distance from the plane affect the E field?

The E field above a plane follows an inverse square law, meaning that it decreases in strength as the distance from the plane increases. This relationship can be described by the formula: E ∝ 1/d2, where d is the distance from the plane.

5. What are some real-life applications of the E field above a plane?

The E field above a plane has many practical applications, such as in the design of electronic devices and equipment. It is also crucial in understanding the behavior of lightning strikes and the formation of thunderstorms. Additionally, the E field above a plane is utilized in the process of electrostatic precipitation, which is used to remove pollutants from industrial exhaust gases.

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