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Hello everyone.
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In this problem we're as to find and show a couple of relations between the electric and magnetic field components of electromagnetic waves.
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Four are given that these waves are standing waves, are the standing wave solutions to the maximal equations.
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Okay, so we have that the y component of the electric field is equal to minus two times e times the sign of k times x times the sign of omens.
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Omega times t and the z component of the magnetic field vector is minus two times b times the cosine of kx times the cosine of omega t okay so we have some basic relationships between these quantities we have that the wave number is 2 pi divided by lambda and we know that lambda is equal to the speed divided by the frequency so then using these two relations we can write the wave a number as 2 pi f over c, which is equal to omega over c, okay, where omega is the angular frequency of the wave.
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We know that the speed is given by one over the root, the square root of epsilon zero and mu zero.
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This is another important relation that we know that we know.
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And then we also know that the amplitudes of the waves are related through the speed of flight or through the speed of the wave in this way.
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Also, here i word out just some basic trig identities.
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So the derivative of the sign function is the cosine function.
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The derivative of the cosine function is the negative of the sign function.
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Okay.
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So for the first part, we have to show that the second derivative, the second partial derivative of the elective.
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Electric field amplitude with respect to x is equal to epsilon 0 times mu 0 times the second partial derivative of the electric field with respect to time.
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So we're going to proceed by looking at both sides of the equation.
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And so looking at the left hand side, we first work out the first derivative, which is minus 2e times k times cosine k x times sine omega t.
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So this is just differentiating this equation here, with respect to x, and then differentiating that equation with respect to x again, or just differentiating this equation with respect to x, we find that the second derivative is 2 times e times k squared, times sine kx times sine omega t.
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So as far as the oscillating functions go, they have gone full circle and they've gone back to their original form.
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But in the meantime, through differentiating by parts twice, we've picked up a minus sign and a k squared factor.
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So now using the fact that k is equal to omega over c, all right, that's what i'm indicating here with this arrow.
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So this right hand or left hand side becomes two times e times omega squared over c squared times sine kx times sine omega t.
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Okay, good.
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Now we're going to look at the right -hand side.
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On the right -hand side, the first derivative of respect to time yields minus 2e omega -syn kx times cosine -omba -t.
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And the second derivative yields 2 times e times omega -squared times sine kx times sine omega -t.
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Now notice that this is just the derivative is.
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We have not taken into account the factor of epsilon 0 times mu 0 on the right -hand side yet.
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So we multiply this second derivative that we've calculated by that.
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Factor to get that we have 2 times e times omega squared times mu 0 times mu 0 times sine k x times sine omega t.
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And now recognizing that epsilon 0 mu 0 is 1 over c squared, we get that this is equivalent to 2 times e times omega squared or c squared times sine k x times sine omega t.
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Now notice that this is actually equal to what we have found on the left hand side...