Choosing Integrating constants for Electric Field

In summary, the electric field at a point on a disk with uniform positive surface charge density is given by: E=kq/r.
  • #1
Jen2114
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Homework Statement


A nonconducting disk of radius R has a uniform positive surface charge density sigma. Find the Electric field at a point along the axis of the disk at a distance x from its center. Assume that x is positive

Homework Equations


E=kq/r

The Attempt at a Solution


I know I'm suppose to find dEX for one ring and then integrate to find the field due to all the rings.
dEx= (k) (2πσrx)dr /(x^2 +r^2) ^3/2
Why should you integrate this component from 0 to R and not R to -R
 
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  • #2
Where exactly is the part of the disk from r=0 to r=-R? That is, where are the negative radius locations?

By the way, it is not an integrating constant as you suggest in the title. This is a definite integral so there is no integrating constant.
 
  • #3
For any future reference,what you mean is called "interval of integration" and not constants.
 
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  • #4
Hi,
sorry you're right I should've said that I don't understand why the limits of integration are 0 to R and not -R to R. The center of the disk is located at (0,0) and so the negative radius is at (0,-R) and the positive is at (0,R). The radius of the first ring I'm integrating is r and so then I have to integrate for the entire disk.
 
  • #5
LittleMrsMonkey said:
For any future reference,what you mean is called "interval of integration" and not constants.
Thank you, I will be much more clear next time
 
  • #6
Think about it logically.You are integrating radially outwards from 0.
 
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  • #7
So dEx=(1/4πε)*((2πσrx)/(x^2+r^2)^3/2)) is the electric field component in the x direction and so when you integrate to obtain the electric field for all the small rings in the disk , you are working your way out towards R, the radius of the entire disk. So that's why you integrate from 0 to R and not -R to R?
 
  • #8
It's easy,see?
You've forgotten the dr in your formula.
 
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  • #9
Ahhh ok I see thanks. Yeah, super clear now. Thanks I'll add the dr. Thanks again!
 
  • #10
You're welcome.I'm studying for an E-M exam right now anyway,so it's good use of my time.
 
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Related to Choosing Integrating constants for Electric Field

1. How do I choose the integrating constant for an electric field?

The integrating constant for an electric field can be chosen based on the boundary conditions of the problem. These conditions include the charge distribution, shape of the conductors, and the potential at different points in the field. It is important to choose a constant that satisfies these conditions and makes the calculations simpler.

2. What is the purpose of an integrating constant in an electric field?

An integrating constant is used to account for the fact that the electric field is a conservative vector field. It is added to the integral of the electric field to ensure that the potential at any point in the field is well-defined and independent of the path chosen for the integral.

3. Can I choose any value for the integrating constant in an electric field?

No, the value of the integrating constant is determined by the boundary conditions of the problem. Choosing an incorrect value can lead to incorrect results and violate the laws of electromagnetism.

4. How does the choice of integrating constant affect the electric field?

The choice of integrating constant does not affect the electric field itself, but it does affect the potential at different points in the field. Different values of the constant can lead to different potential values, but the electric field and its behavior remain the same.

5. Can the integrating constant be changed after solving for the electric field?

No, once the integrating constant has been chosen and the electric field has been solved for, it cannot be changed. To incorporate a different constant, the calculations for the electric field would have to be redone with the new constant in mind.

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