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

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The discussion centers on verifying that the equation E = E(max)cos(kx-wt) is a solution to Maxwell's derived equation, specifically the wave equation represented as ((d^2)E/dx^2) = ε₀μ₀(d²E/dt²). Participants emphasize the importance of substituting k = 2π/λ and ω = 2πf, and using the relationship c = 1/√(ε₀μ₀) to connect the wave equation's components. The method involves performing partial differentiation on E(x,t) and equating the results to confirm the solution. The conversation highlights the necessity of understanding the mathematical operations involved in the verification process.

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maon
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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
 
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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.
 
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?
 
The only relation you need to use is \omega/k=c=\frac{1}{\sqrt{\epsilon_0 \mu_0}}.

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

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

So does E(x,t) verify the equation?
 
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.
 
Well, isn't that what the wave-equation says?

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

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.
 
Thanks, I think I got it. So you derivative for B and E and solve both sides. I couldn't get the question, it was worded kinda weird. Thanks for your help.
 
What answer did you get?
 
maon said:
Thanks, I think I got it. So you derivative 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.")
 
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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.
 

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