What is the Solution to Gauss's Law for a Charged Plate Capacitor?

In summary: They focus on the application of physics concepts to a real-world scenario. This should give you a better idea of how to approach the problem, and help you to develop a intuition for how physics works.In summary, the student is having difficulty solving a homework problem that asks for the distance between two plates. The extraneous information given about the distance is confusing and does not help the student to solve the problem. The student is recommended to try the exercises from Purcell's book that focus on applying physics concepts to a real-world scenario.
  • #1
exitwound
292
1

Homework Statement



problem.jpg


Homework Equations



[tex]\Phi = \int{\vec E \cdot d\vec A}= \frac{q_{enc}}{\epsilon_0}[/tex]


The Attempt at a Solution



The positive plate is drawn to the left of the negative plate. If we draw a Gaussian surface around the inner edge of of the positive plate, we can see that E and A are both pointing in the +x direction. Therefore: [itex]\epsilon_0EA=q_{enc}[/itex]

However, I have no idea where to go from here. Certainly I can't just solve for Qenc if they give me distances between plates in the problem.

How do I continue?
 
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  • #2
exitwound said:
Certainly I can't just solve for Qenc if they give me distances between plates in the problem.
Says who? :wink:
 
  • #3
This isn't just a teacher's problem tho. Comes from a textbook. Why would they give you EXTRA information? I've rarely ever come across something like that.

What purpose does giving the distance serve? I know that the Efield is independent of distance when talking about infinite plates. And since we're negating the edging in this problem, that's what we're dealing with. But giving extraneous numbers leads to confusion.

So if I just plopped in:

(8.85e-12)(60)(1.2) I should get the right answer.
 
  • #4
Sure, it's a bit odd to give extraneous information, but it's not out of the question. It's up to you to realize that it's extraneous. (Are there other parts to this problem?)

The distance is not entirely extraneous. What if the distance was 4.8 m instead of 4.8 cm?
 
  • #5
Understood about the distances between the plates. There are no other parts to this question. I don't believe textbooks are out there to swindle us, or cause us pain. Normally, at least in all textbooks that I've encountered, the problem gives you exactly what you need to solve it. Here's A, find B. In this way, they teach along constant lines. Certainly, there are multifaceted questions that often ask "does this make sense?" and "what does X rely on?" But when problems typically ask for absolute answers, they normally feed you what you need, and nothing more.

So far, I'm really really unimpressed with the problem sets in this book.

Fundamentals of Physics, 8th Ed, Jearl Walker (halliday/resnick)
 
  • #6
exitwound said:
So far, I'm really really unimpressed with the problem sets in this book.

Fundamentals of Physics, 8th Ed, Jearl Walker (halliday/resnick)

Try the "few" (compared to Halliday's) exercises of Purcell's book (Berkeley Physics course, volume 2).
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the distribution of electric charge to the resulting 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.

2. How is Gauss's Law used in physics?

Gauss's Law is used to calculate the electric field in situations of symmetry, where the electric field can be assumed to be constant over a closed surface. It is also used to determine the charge distribution in a given system.

3. What is the mathematical formula for Gauss's Law?

The mathematical formula for Gauss's Law is ∮E⃗ · dA⃗ = Q/ε0, where ∮E⃗ · dA⃗ represents the electric flux through a closed surface, Q is the total charge enclosed by the surface, and ε0 is the permittivity of free space.

4. What is the importance of Gauss's Law in electricity and magnetism?

Gauss's Law is important in electricity and magnetism because it allows us to understand and predict the behavior of electric fields and their relationship to electric charges. It also helps us calculate the electric field in various situations, making it a fundamental tool in solving problems in electromagnetism.

5. Can Gauss's Law be applied to magnetic fields?

No, Gauss's Law only applies to electric fields. However, there is a similar law for magnetic fields called Gauss's Law for Magnetism, which states that the magnetic flux through a closed surface is always zero.

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