Calculate grad X E Am I on the right track?

[frozenguy]

I only did the first two (out of three) crosses because I want to check that I'm on the right track. It doesn't seem like I am because I have come up with 0i, 0j and I pretty much know I will come up with 0k

 

[RoyalCat]

On a side note, \vec \nabla \times \vec E is read "del cross E", or "rotor of E/curl of E." The Greek letter Nabla in this context is usually called "the del operator."

The gradient of a scalar function, \phi is denoted by grad \phi or \nabla \phi, and is just one use for the del operator.

The divergence of a vector function, let's take E for now, is denoted by: div \vec E or \vec \nabla \cdot \vec E

Now that that's out of the way, 0 makes sense. The curl of a conservative field is 0! The electric field you've described here, is that of a point charge q at rest. Such a field is conservative, and thus, we should expect \vec \nabla \times \vec E=0, which you have demonstrated exactly. :)

If you've already learned Maxwell's Equations in their differential form, then you should know that \vec \nabla \times \vec E=-\frac{\partial\vec B}{\partial t} and since in this case you're dealing with a single point charge at rest, there's no time variation of the magnetic field anywhere, again you should expect \vec \nabla \times \vec E=0

 

[frozenguy]

On a side note, \vec \nabla \times \vec E is read "del cross E", or "rotor of E/curl of E." The Greek letter Nabla in this context is usually called "the del operator."

The gradient of a scalar function, \phi is denoted by grad \phi or \nabla \phi, and is just one use for the del operator.

The divergence of a vector function, let's take E for now, is denoted by: div \vec E or \vec \nabla \cdot \vec E

Now that that's out of the way, 0 makes sense. The curl of a conservative field is 0! The electric field you've described here, is that of a point charge q at rest. Such a field is conservative, and thus, we should expect \vec \nabla \times \vec E=0, which you have demonstrated exactly. :)

If you've already learned Maxwell's Equations in their differential form, then you should know that \vec \nabla \times \vec E=-\frac{\partial\vec B}{\partial t} and since in this case you're dealing with a single point charge at rest, there's no time variation of the magnetic field anywhere, again you should expect \vec \nabla \times \vec E=0

Hey thanks!
Hmm ok so it should equal 0 because the particle is at rest, so the field is conservative? What is the del operator? As in what is it exactly am I crossing with E besides partial derivatives.

 

[RoyalCat]

Hey thanks!
Hmm ok so it should equal 0 because the particle is at rest, so the field is conservative? What is the del operator? As in what is it exactly am I crossing with E besides partial derivatives.

It is often useful to consider the del operator, \nabla as a vector whose components are the partial derivative operators:

\vec \nabla =( \frac{\partial}{\partial x}) \hat x+( \frac{\partial}{\partial y}) \hat y+( \frac{\partial}{\partial z}) \hat z

Now we can look at the 3 vector operations carried out using the del operator.

grad (\phi) = \vec \nabla \phi = ( \frac{\partial \phi}{\partial x}) \hat x+( \frac{\partial \phi}{\partial y}) \hat y+( \frac{\partial \phi}{\partial z}) \hat z

div (\vec v)=\vec \nabla \cdot \vec v = (\frac{\partial \vec v}{\partial x})\cdot \hat x+(\frac{\partial \vec v}{\partial y})\cdot \hat y+(\frac{\partial \vec v}{\partial z})\cdot \hat z
Notice the dot product between the partial derivatives and the unit vectors(!), the divergence of a vector field is a scalar function, and not a vector one.

And as for curl (\vec v) = \vec \nabla \times \vec v I won't write that out since I'm lazy, and you've shown you can use the pseudo-determinant to take the cross product.
What you posted in your solution holds for any two vectors whose cross product you wish to find, just replace the components of the del operator, the partial derivatives, with the components of any other vector.

 

[frozenguy]

It is often useful to consider the del operator, \nabla as a vector whose components are the partial derivative operators:

\vec \nabla =( \frac{\partial}{\partial x}) \hat x+( \frac{\partial}{\partial y}) \hat y+( \frac{\partial}{\partial z}) \hat z

Now we can look at the 3 vector operations carried out using the del operator.

grad (\phi) = \vec \nabla \phi = ( \frac{\partial \phi}{\partial x}) \hat x+( \frac{\partial \phi}{\partial y}) \hat y+( \frac{\partial \phi}{\partial z}) \hat z

div (\vec v)=\vec \nabla \cdot \vec v = (\frac{\partial \vec v}{\partial x})\cdot \hat x+(\frac{\partial \vec v}{\partial y})\cdot \hat y+(\frac{\partial \vec v}{\partial z})\cdot \hat z
Notice the dot product between the partial derivatives and the unit vectors(!), the divergence of a vector field is a scalar function, and not a vector one.

And as for curl (\vec v) = \vec \nabla \times \vec v I won't write that out since I'm lazy, and you've shown you can use the pseudo-determinant to take the cross product.
What you posted in your solution holds for any two vectors whose cross product you wish to find, just replace the components of the del operator, the partial derivatives, with the components of any other vector.

Thank you so much! Really cleared a lot up.