Electric Field at a Point Near a Charged Rectangular Plate

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field at a point near a charged rectangular plate using the equation dE = 1/(4πε₀) dq /r². The user attempts to derive the electric field vector by considering a uniformly distributed charge over the plate, parametrizing the plate's dimensions, and applying symmetry to cancel out components. However, they encounter difficulties in integrating the resulting expressions and seek further resources for a comprehensive solution, particularly for points not assuming an infinite plane.

PREREQUISITES
  • Understanding of electric field concepts and equations, specifically dE = 1/(4πε₀) dq /r²
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of vector calculus, including gradient and unit normal vectors
  • Basic principles of electrostatics and charge distribution
NEXT STEPS
  • Research methods for calculating electric fields from charged surfaces, focusing on finite charge distributions
  • Study the application of symmetry in electrostatics to simplify electric field calculations
  • Explore integration techniques for multi-variable functions in the context of electric fields
  • Learn about numerical methods or software tools like Wolfram Alpha for solving complex integrals
USEFUL FOR

Students and professionals in physics, particularly those studying electrostatics, electrical engineering, or anyone involved in calculating electric fields from charged surfaces.

Harrisonized
Messages
206
Reaction score
0

Homework Statement



Given an arbitrary curve or surface with a total charge of Q, find the vector equation for the electric field at any point in space.

Homework Equations



dE = 1/(4πε₀) dq /r2

The Attempt at a Solution



Problem 1

Take the unit circle on the plane, for example. Find the vector equation for the electric field at a point (x0,y0).

The unit circle on the plane can be parametrized by
x = r cos θ
y = r sin θ

or alternatively,

r = √(x2+y2)

I know that the electric field lines generated are normal to such a curve. Therefore, if I take the gradient:

∇r(x,y) = 1/r*(x,y)

I get the unit normal vector. The only way I know how to do this problem is by letting Q be concentrated in the middle of the circle to obtain

E = 1/(4πε₀) Q /r2 * 1/r*(x,y)

Unfortunately, this is wrong for the interior of the sphere.

I'm not really sure what to do. I've looked in so many textbooks, and none of them tell how to actually find the vector field E. All the sources I found only concern themselves with finding |E| at a point specified from the "object" of charge, and most of them just use symmetry to reduce the problems to pointlessness.

Any help is appreciated. If possible, please link me to some source that provides the method of finding such fields.

Problem 2

Find the electric field at a point a perpendicular distance a away from center a rectangular plate of uniformly distributed charge. The rectangular plate has dimensions m*n and no thickness.

2. Homework Equations

dE = 1/(4πε₀) dq /r2

The Attempt at a Solution



Let dq = σ dx dy, where σ = Q/(mn), Q is the total charge of the plate.

Let the center of the plate be the origin of x and y. Then:
r=√(x2+y2+a2)

By symmetry, the x and y components along the direction of the plane cancel out. Let θ represent the angle between a and rx and φ represent the angle between a and ry. Then:

dE = 1/(4πε₀) dq /r2
dE = 1/(4πε₀) σ dx dy /r2 * cos θ * cos φ
dE = 1/(4πε₀) σ dx dy /(x2+y2+a2) * a/√(a2+x2) * a/√(a2+y2)
Integrate...
... and...
Wolfram Alpha: (No result found in standard mathematical functions.)

:(

Also, how do you do this for any point (without assuming an infinite plane)?
 
Last edited:
Physics news on Phys.org
I found the solution to Problem 1.

http://www.physics.buffalo.edu/~sen/documents/field_by_charged_ring.pdf

Anyone want to help me on Problem 2?
 
Last edited by a moderator:

Similar threads

  • · Replies 79 ·
3
Replies
79
Views
10K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
9K
Replies
4
Views
1K