Question about cross and dot product

In summary, the conversation is about proving the equations \nu*E=0 and \nu x E=-dB/dt using the electric field E and the Maxwell equations. The person is confused about the steps taken in the solution, such as using 1/s d/ds(s E) instead of d/ds(E) and the appearance of z^-1/s in the cross product. They also mention the boundary conditions of having E parallel = 0 and B perpendicular = 0 and using the dot product to show that E satisfies the Maxwell equations.
  • #1
leonne
191
0

Homework Statement



Note: [tex]\nu[/tex] is del could not find it...

I need to prove the [tex]\nu[/tex]*E=0
and [tex]\nu[/tex] x E=-dB/dt

Homework Equations


E(s[tex]\phi[/tex]zt)=(Acos(Kz=wt)/s)s^

The Attempt at a Solution


so for the first one [tex]\nu[/tex]*E=0 I thought it would be
d/ds(E) but what they did was 1/s d/ds(s E) no idea why they did it. Is it because its S^? I thought i remember doing something like this.
for the cross product
d/ds d/d[tex]\phi[/tex] d/dz
E 0 0

...I got the right answer but they got 0+dE/dZ [tex]\phi[/tex]^-1/s dE/d[tex]\phi[/tex] z^
no idea how they got the z^ i thought it was 0, well it does =0 but where did the -1/s come from?
thanks for the help[STRIKE][STRIKE][/STRIKE][/STRIKE]
 
Last edited:
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  • #2
try writing all youyr code in a single tex banner & to write del (\nabla), so you want to show:
[tex]\nabla \bullet E = 0 [/tex]

how about directly applying the product? I can;t really understand teh form of your electric field - mayeb you can re-write it?
 
  • #3
hmm can't edit my post
so need to show that
[tex]\nabla[/tex] [tex]\bullet[/tex] E=0
[tex]\nabla[/tex] [tex]\times[/tex] E= -dB/dT
Well its a wave function
And i need to show that E satisfy Maxwell equation and the boundary conditions.
boundary condition E parallel =0 and B perpendicular=0
lets say E was x^ for the dot product
(d/dx+d/dy+d/dz)(E x^+0+0) so i would have dE/dx am i wrong?
 

Related to Question about cross and dot product

1. What is the difference between the cross product and the dot product?

The cross product and the dot product are both mathematical operations that involve two vectors. The main difference between them is that the cross product results in a vector, while the dot product results in a scalar. The cross product is also known as the vector product, while the dot product is also known as the scalar product.

2. How do you calculate the cross product?

The cross product of two vectors, a and b, is given by the formula a x b = (|a| * |b| * sinθ) * n, where |a| and |b| are the magnitudes of the vectors, θ is the angle between them, and n is a unit vector perpendicular to both a and b. Another way to calculate the cross product is by using the determinant of a 3x3 matrix with the components of a and b as its rows.

3. What is the physical significance of the cross product?

The cross product has several physical interpretations. It can represent the magnitude of torque (rotational force) when a force is applied at a distance from a fixed point. It can also represent the magnitude of the magnetic field created by a current-carrying wire. In physics, the cross product is often used to describe the direction and magnitude of rotational or circular motion.

4. Can the cross product be applied to more than two vectors?

No, the cross product is only defined for two vectors in three-dimensional space. It is not possible to calculate the cross product of more than two vectors.

5. How is the dot product used in vector projections?

The dot product can be used to calculate the length of a projection of one vector onto another. This is useful in physics and engineering for calculating the amount of force acting on an object in a certain direction. The dot product can also be used to determine the angle between two vectors, which can be helpful in understanding the relationship between different quantities in a system.

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