Electric Field w/ Continuous Charge Dist.

In summary, you are trying to find the magnitude and direction of the electric field at the origin. You use the following expressions: (k_e)(lambda) int{x_otoinf} x^-2 dx, (k_e)(lambda) int{x_otoinf} - 1 / x, and (k_e)(lambda) [(-1 / x_o) - (1 / inf)] cos 0. You find that (-k_e lambda / x_o) + (-k_e lambda / inf) i and (-k_e)(lambda) int{x_otoinf} x^-2 dx = k_e int lambda dx / x^2. You then use this
  • #1
oooride
34
0
Okay I'm really confused with continuous charge distributions, so I'm totally stuck...


A continuous line of charge lies along the x axis, extending from x=+x_0 to positive infinity. The line carries a uniform linear charge density lambda_0. What are the magnitude and direction of the electric field at the origin?


I have no idea where to approach or how to setup this problem up really.. What I tried to do is this..

Since, lambda = Q / L = dq / dx

E = int dE

= int k_e dq / r^2

= k_e int lambda dx / x^2

And then I'm completely stuck... Do I need to find lambda or just integrate? or...?


Any help is greatly appreciated. :smile:
Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
You're on the right track. Lambda is a constant, so put it out in front along with ke & integrate, (but first, determine your limits of integration).
 
  • #3
Thanks for the quick response. I'm also confused on what my limits of integration are.. Do I go from {x_o to infinity}, or {0 to x_o}, or ...?

Okay so to continue on,

(k_e)(lambda) int{x_otoinf} x^-2 dx

(k_e)(lambda) int{x_otoinf} - 1 / x

(k_e)(lambda) [(-1 / x_o) - (1 / inf)] cos 0

[(-k_e lambda / x_o) + (-k_e lambda / inf)] i


Thanks for the help.:smile:
 
Last edited:
  • #4
Good.

Now, do you understand why it is {x_otoinf} or was that a lucky guess?

Where did cos 0 come from?

and what are you planning to do with (-k_e lambda / inf)?
 
  • #5
Okay well the limits were partially a guess.. But I figured from x_o to infinity because it is a continuous line of charge along the x-axis but I wasn't sure because like you said I don't know what to do with (-k_e)(lambda)/inf.


I figured cos 0 because I thought it was a vector that lies along the x-axis but has no angle and no y component(j). So, cos 0 = 1 and the answer would have +i given in the notation. Is this wrong?
 
  • #6
You're right about the limits. When you compute this integral, you are *summing* the electrical fields arising from the continuous line of tiny charges dq which runs from x_0 to infinity -- the location of the charges is what determines the limits of integration.

I would say the direction of the field is -i. Since it wasn't specified that the charge is negative, I would assume it is positive.
The field direction is away from a positive charge, and the origin is located to the left of the line of charge. So the field at the origin is directed towards the left.

As to the cosine, yes, cos 0 = 1 and yes, this field is along the x-axis and yes, there is no y component. It's just odd that the cosine just pops up out of nowhere. You're not doing a dot product here, and you didn't have any trig functions in the expression you are integrating. So it didn't arise out of anything in your computation and it doesn't really serve any purpose. You can explain your reasoning behind the direction of the field just as well without it, and in fact you'd have to give the same explanation to justify the cosine itself. So I'd say it doesn't belong there.

And as to (-k_e)(lambda)/inf
what is the value of x/ ∞ (where x = any finite number)?
 
  • #7
could be zero. I'm confused on how that effects lambda though.
 
Last edited:
  • #8
My point is just that
[tex]-\frac{k_e \lambda}{\infty} = 0 [/tex]
so that entire term drops out, and the final result is just
[tex] E = - \frac{k_e \lambda}{x_0} \bold{i} [/tex]
 
  • #9
Ah okay.. I understand now.

Thanks for the help! :smile:
 

What is an electric field with continuous charge distribution?

An electric field with continuous charge distribution refers to the presence of a continuous distribution of electric charges that produce an electric field in the surrounding space. This is in contrast to discrete charges, which are individual point charges that create an electric field.

How is the strength of an electric field with continuous charge distribution determined?

The strength of an electric field with continuous charge distribution is determined by the magnitude and distribution of the charges present. The electric field strength at a specific point is given by the Coulomb's law equation, which takes into account the distance between the charges and their respective magnitudes.

What is the relationship between electric field and electric potential in a continuous charge distribution?

The electric field and electric potential are closely related in a continuous charge distribution. The electric potential is a scalar quantity that describes the electric potential energy per unit charge at a given point. It is related to the electric field by the equation E = -∇V, where E is the electric field and V is the electric potential.

How does the presence of a dielectric material affect an electric field with continuous charge distribution?

A dielectric material is a material that can be polarized by an electric field, creating an induced dipole moment. This polarization can alter the electric field in the material, resulting in a weaker overall electric field. However, the electric field outside the dielectric material remains unchanged.

What are some real-world applications of electric fields with continuous charge distribution?

Electric fields with continuous charge distribution have various applications in our daily lives. They are used in electronic devices such as capacitors, transistors, and displays. They are also essential in medical imaging techniques such as MRI and PET scans, as well as in particle accelerators used in scientific research. Other applications include electrostatic precipitators in air purifiers and electrostatic spraying in agriculture.

Similar threads

  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
743
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
28
Views
557
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
696
Back
Top