Is E = E(max)cos(kx-wt) a Solution to Maxwell's Derived Equation?

In summary, the author is trying to verify that E = E(max)cos(kx-wt) is a solution to Maxwell's derived equation. They substitute k = 2pi/lambda, w = 2pi(f), and set 2pi(f) as 2pi/lamda x c. They then take 2pi/lambda out and times cos equals 1/lambda. They are still lost, so a friend helps them out. They solve for E(x,t) by doing partial differentiation for E(x,t) = Emax(cos(kx-wt)).
  • #1
maon
5
0
Hi,
I'm new here. I have a question. How do you verify that: E = E(max)cos(kx-wt) is a solution to Maxwell's derived equation:
((d^2)E/dx^2) = e(epsilon nought)u(permitivity of free space) x (d^2)E/dx^2. Thanks.
What I first did was to substitute k = 2pi/lambda, w = 2pi(f). Then I set 2pi(f) as 2pi/lamda x c. Taking 2pi/lambda out and times cos equals 1/lambda. Now I'm stuck.
Mat
 
Last edited:
Physics news on Phys.org
  • #2
To verify the function satisfied the wave equation, you simply put in there. Or look what the left side of the wave equation is for the given E, then calculate the right side and see if they are equal.
 
  • #3
I'm still confused in terms of how to get the sides to equal each other. How do I eliminate the minus and the cos?
 
  • #4
The only relation you need to use is [itex]\omega/k=c=\frac{1}{\sqrt{\epsilon_0 \mu_0}}[/itex].

So you have [itex]E(x,t) = E_{max}\cos(kx-wt)[/itex].

Now what's [itex]\frac{\partial^2}{\partial x^2}E(x,t)[/itex] and what's [itex]\frac{\partial^2}{\partial t^2}E(x,t)[/itex]?

So does E(x,t) verify the equation?
 
  • #5
I'm sorry I'm a bit lost here. So you do partial differentiation for E(x,t) = Emax(cos(kx-wt)) for x and t? Thank you for your patience.
 
  • #6
Well, isn't that what the wave-equation says?

[tex]\frac{\partial^2}{\partial x^2}E(x,t)=\epsilon_0 \mu_0 \frac{\partial^2}{\partial t^2}E(x,t)[/tex]

What exactly is it you don't understand? Is it the question itself or the way to go about solving the problem? My guess is that you don't understand the question.
 
  • #7
Thanks, I think I got it. So you derivitive for B and E and solve both sides. I couldn't get the question, it was worded kinda weird. Thanks for your help.
 
  • #8
What answer did you get?
 
  • #9
maon said:
Thanks, I think I got it. So you derivitive for B and E and solve both sides. I couldn't get the question, it was worded kinda weird. Thanks for your help.
Actually, I would think you were the one with the "weird" wording. I have no idea what "solve both sides" means! I know how to solve equations and I even know how to solve problems in general but I don't know what is meant by solving a "side" of an equation!


Suppose you had a problem that asked to show that x= 3 is a solution to x2- 5x= -6. What would you do? (I hope you would not say "solve the equation.")
 
  • #10
I just did a dervative twice for E = Emaxcos(kx-wt) and for B = Bmaxcos(kx-wt) in terms of x and t. Then I equaled the double derivative of E = Emaxcos(kx-wt) in terms of x with terms of t and solve to get light = light. Same thing for B equation too.
 

FAQ: Is E = E(max)cos(kx-wt) a Solution to Maxwell's Derived Equation?

What are Maxwell's equations?

Maxwell's equations are a set of four fundamental equations that describe the behavior of electromagnetic fields. They were first published by James Clerk Maxwell in the 1860s and are considered one of the cornerstones of classical electromagnetism.

What are the four equations in Maxwell's equations?

The four equations are Gauss's law, which relates the electric field to the electric charge; Gauss's law for magnetism, which relates the magnetic field to the magnetic charge; Faraday's law of induction, which describes how a changing magnetic field creates an electric field; and Ampere's law, which relates the magnetic field to the electric current.

Why are Maxwell's equations important?

Maxwell's equations are important because they provide a complete description of the behavior of electromagnetic fields, including how they are created and how they interact with matter. They have been extensively tested and are the basis for many technological applications, such as wireless communication and electricity generation.

How do Maxwell's equations relate to light?

Maxwell's equations were originally developed to describe the behavior of electromagnetic fields, which include both electric and magnetic fields. Light is a type of electromagnetic radiation, so Maxwell's equations also describe the behavior of light. This led to the unification of electricity, magnetism, and optics into the field of electromagnetism.

Are Maxwell's equations still relevant today?

Yes, Maxwell's equations are still relevant today and are widely used in many fields of science and engineering. They have been confirmed by numerous experiments and are considered one of the most accurate and fundamental theories in physics. They also continue to be a basis for new discoveries and technological advancements.

Similar threads

Replies
6
Views
3K
Replies
4
Views
1K
Replies
21
Views
4K
Replies
4
Views
2K
Replies
10
Views
1K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
3
Views
1K
Back
Top