Recent content by maon
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Is E = E(max)cos(kx-wt) a Solution to Maxwell's Derived Equation?
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.- maon
- Post #10
- Forum: Introductory Physics Homework Help
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Is E = E(max)cos(kx-wt) a Solution to Maxwell's Derived Equation?
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.- maon
- Post #7
- Forum: Introductory Physics Homework Help
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Is E = E(max)cos(kx-wt) a Solution to Maxwell's Derived 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.- maon
- Post #5
- Forum: Introductory Physics Homework Help
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Is E = E(max)cos(kx-wt) a Solution to Maxwell's Derived Equation?
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?- maon
- Post #3
- Forum: Introductory Physics Homework Help
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Is E = E(max)cos(kx-wt) a Solution to Maxwell's Derived Equation?
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)...- maon
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- Maxwell's equation
- Replies: 9
- Forum: Introductory Physics Homework Help