How Does Charge Distribution Affect Electric Field and Work Done?

In summary, Gauss' law tells us that the electric field produced by a charge is proportional to the charge's mass and the distance from the charge.
  • #1
uchicago2012
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0

Homework Statement


Charge is uniformly distributed along the x-axis with density ß. Use Gauss' Law to find the electric field it produces, and use this to calculate the work done on a charge Q that moves along the y-axis from y = a to y = b.

Homework Equations


Gauss' Law: e0 * electric flux = qencl
where e0 is epsilon0

The Attempt at a Solution


I thought dq = Bdx
q = B integral (x)
= (Bx2)/2
Then if we say EA = qencl/eo
where E is the electric field and A is the area
The areas wind up canceling and I got
E = B/e0
But I'm not sure that's right at all. Particularly since I can't see how I use this electric field to calculate the work done on a charge Q that moves along the x axis. Wouldn't I have to know if the charges were positive or negative? Because presumably opposite charges would cause the charge Q to speed up as it approaches the x-axis while same would cause it to slow down... or am I on the wrong track completely?
 
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  • #2
uchicago2012 said:
I thought dq = Bdx
That sounds reasonable to me. :approve:
q = B integral (x)
Wait, where did the x come from? And what happened to your dx? :wink:
 
  • #3
Oh, I see. So I've found the charge, then Q = Bx. But then how do I go about finding the electric field? I can find the flux by dividing by e0 of course. But the flux = integral (flux * dA) which I'm not sure applies to this problem. Or I can say flux = E*A, which I've seen my book randomly do but I don't know if it applies in this case. Is there a general way of trying to find the electric field? Like once you know the flux, shouldn't you theoretically be able to calculate the electric field somehow?
 
  • #4
uchicago2012 said:
Oh, I see. So I've found the charge, then Q = Bx.
That's better. :approve:
But then how do I go about finding the electric field? I can find the flux by dividing by e0 of course. But the flux = integral (flux * dA) which I'm not sure applies to this problem. Or I can say flux = E*A, which I've seen my book randomly do but I don't know if it applies in this case. Is there a general way of trying to find the electric field? Like once you know the flux, shouldn't you theoretically be able to calculate the electric field somehow?
Look for symmetry. You need to mentally find some sort of surface, called a Gaussian surface, such that the E·dA is constant over any given section of that surface (and it's okay to break up the closed integral into sections if you have to). Look for the situations where the direction dA is parallel to E for all dA of that section of the closed surface (dA points in a direction normal to the surface). If you can find such a Gaussian surface, then

[tex] \int_S \vec E \cdot \vec {dA} = \int_S EdA = E \int_S dA [/tex]

over that section of the surface. (Remember the above situation does not apply for everything -- you need to have a Gaussian surface such that E·dA is constant.) Note how in this situation, the dot product is merely the multiplication of the the magnitudes, and now E becomes a simple, constant scalar and can be pulled out from under the integral.

By the way, there are only a few situations that you can use this method to find E. You can use it for problems like this one where there is cylindrical symmetry, and the charged line or charged cylinder is infinitely long. You can also use for problems with spherical symmetry. And finally, you can use it for infinitely large planes. But that's about it.

Don't get me wrong, Gauss' law is always true in all situations (classical electrodynamics, that is). And if you happened to already know E for all space, you can always use Gauss' law to find the charge enclosed within any closed surface. But the above situations are the only ones that you will use it to find E.
 
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  • #5



You are on the right track with using Gauss' Law to find the electric field produced by the charge distribution. However, it is important to note that the charge density, ß, is not given in the problem statement and thus cannot be used in your calculation. Instead, you can use the total charge, Q, and the length of the x-axis to find the charge density, which can then be used in your calculation for the electric field.

Once you have the electric field, you can calculate the work done on a charge Q moving along the y-axis by using the equation W = QEd, where d is the distance traveled along the y-axis (from y = a to y = b). This work done will depend on the direction of the charge Q and the direction of the electric field, so it is important to consider the sign of the charge Q in your calculation.

In summary, to find the electric field produced by the charge distribution, you can use Gauss' Law and the total charge Q. To calculate the work done on a charge Q moving along the y-axis, you can use the equation W = QEd, where E is the electric field found using Gauss' Law. The direction of the charge Q and the electric field should be taken into account in your calculation.
 

What is Gauss' Law?

Gauss' Law is a fundamental law in physics that describes the relationship between electric charge and the electric field. It states that the electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space. In simpler terms, it explains how electric charges create electric fields.

How is Gauss' Law different from Coulomb's Law?

While both laws deal with electric charges and fields, they have different applications. Coulomb's Law is used to calculate the electric force between two stationary charges, while Gauss' Law is used to calculate the electric field created by a distribution of charges. Additionally, Gauss' Law is more useful for symmetrical charge distributions, while Coulomb's Law can be used for any configuration of charges.

Can Gauss' Law be applied to both open and closed surfaces?

No, Gauss' Law is only applicable to closed surfaces. This is because the law relates the electric flux through a closed surface to the total enclosed charge. For open surfaces, the electric flux is not well-defined as it can escape through the opening.

What is the significance of the Gaussian surface used in Gauss' Law?

The Gaussian surface is an imaginary shape that is chosen in such a way that it simplifies the calculation of the electric field. By choosing a symmetrical shape, the electric field can be assumed to be constant over the surface, making the calculations easier.

How is Gauss' Law used in real-life applications?

Gauss' Law has many practical applications, including in circuit analysis, electric field mapping, and understanding the behavior of electric charges in various materials. It is also used in the development of electronic devices, such as capacitors and transistors. Additionally, it is an essential tool in the study of electromagnetism and the design of antennas and other communication devices.

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