Electrostatic fields- What defines an electrostatic field?

Click For Summary
An electrostatic field is defined by having no curl and a divergence that indicates the presence of static charges. The discussion revolves around analyzing two given equations to determine if they represent an electrostatic field by calculating their curl and divergence. It is clarified that a stationary charge would produce a field with zero curl and a divergence that is consistent in all directions. The relationship between divergence and charge density is emphasized through Gauss's Law, which connects the divergence of the electric field to the charge distribution. Understanding these properties helps in identifying whether a field is electrostatic.
KiNGGeexD
Messages
317
Reaction score
1
I have a question about electrostatic fields!

I was given these two equationsE= k [xy xˆ + 2yz yˆ + 3xz zˆ]

E= k[y^2 xˆ+ (2xy+ z^2)+ 2yzzˆ]

And asked to determine if either could be an electrostatic field, I'm not asking how to solve this problem but I'm not sure what defines an electrostatic field!

I have heard vector and scalar fields being mentioned and I know that curl and divergence pertain to vector fields and gradient to scalar fields:)

My initial thought was to calculate divergence and curl and draw conclusions from there?

I've combed through passages on my textbook and notes and can't find an explicit answer as to what defines an electrostatic field (other than static point charges)!Any help would be great :)
 
Physics news on Phys.org
Oh I also missed out my "y" unit vector in the second expression, apologies
 
Try applying Faraday's law.
 
I thought that but it hasnt been mentioned in lectures so I was hesitant
 
I'm assuming manipulation of the functions is where the answer is, so is there no relationship between curl, divergence and gradient and electrostatics?
 
Just think of what a static field means. Think of what it implies about the direction of the field lines. If I have a stationary charge (static), what would I expect its divergence to look like? What about its curl?
 
I would expect curl to be zero?

And divergence to be positive or well rather the same in all directions
 
KiNGGeexD said:
I would expect curl to be zero?

And divergence to be positive or well rather the same in all directions

Correct, now go one step further and think about what the divergence implies about the physical charge in space, and what you would expect to happen when you move the charge. Compare what you think to Maxwell's Equations (Gauss' Law and Faraday's Law)
 
I know if the charge were moved the field would remain unchanged
 
  • #10
Are you sure about that?
 
  • #11
Not anymore haha! So would if charges were moved the field could potentially develop a curl and differ in divergence
 
  • #12
KiNGGeexD said:
Not anymore haha! So would if charges were moved the field could potentially develop a curl and differ in divergence

Well the divergence would remain unchanged, but yes, a curl would develop. This curl is a time-varying magnetic field. Charges in motion produce time-varying magnetic fields!
 
  • #13
That makes complete sense!

So to I would know if I had an electrostatic field if it had no curl, and if it's divergence was in the same direction for each component?

Would gradient not then be in the radial direction of the field as well in this case?
 
  • #14
KiNGGeexD said:
I was given these two equations

E= k [xy xˆ + 2yz yˆ + 3xz zˆ]

E= k[y^2 xˆ+ (2xy+ z^2)+ 2yzzˆ]

And asked to determine if either could be an electrostatic field

This is Problem 2.20 in Griffiths, "Introduction to Electrodynamics" (3rd ed.).

I've combed through passages on my textbook

If you're using Griffiths, flip back to section 2.2.4, "The Curl of E". :wink:
 
  • #15
It just so happens I am using Griffiths haha!

Thanks a bunch friend! I appreciate it
 
  • #16
KiNGGeexD said:
and if it's divergence was in the same direction for each component?

Divergence (##\vec \nabla \cdot \vec E##) is a scalar. It doesn't have a direction. As for whether the divergence needs to be anything in particular, what is the divergence related to, via Gauss's Law?
 
  • #17
Divergence is related to Gauss law asΔ.E= ρ/εo
 
  • #18
Whoops that was supposed to be del, not delta
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
6K
  • · Replies 4 ·
Replies
4
Views
1K