Show that the given electric field is a plane wave

Click For Summary
The discussion focuses on demonstrating that a given electric field represents a plane wave. It begins with the definition of a wavefront and the manipulation of the cosine argument to establish a relationship involving position and time. A participant identifies an error in the rearrangement of the equation but acknowledges that it resembles the scalar equation of a plane. The conversation confirms that the derived equation indeed indicates a plane wave, emphasizing the simplicity of the conclusion. Overall, the participants agree on the relationship between the equation and the characteristics of a plane wave.
Blanchdog
Messages
56
Reaction score
22
Homework Statement
Show that each wavefront of the electric field forms a plane
Relevant Equations
## E(r, t) = E_0 \text{cos}(k(\hat{u} \cdot r - c t) + \phi)##
A wavefront is defined as a surface in space where the argument of the cosine has a constant value. So I set the argument of the cosine to an arbitrary constant s.

## k(\hat{u} \cdot r - c t) + \phi = s ##

The positional information is is in r, so I rearrange the equation to be

## \hat{u} \cdot r = \frac s k + ct + \phi = \text{const}##
## u_x x + u_y y + u_z z = \text{const} ##

And that's where I'm stuck.
 
Physics news on Phys.org
You made a small error in the rearrangement, but once corrected the last line is almost right; that's the general equation of a plane of normal ##\hat{n}##, but ##\hat{u} \cdot r## depends on ##t## so isn't constant in time (that's why the plane translates).
 
ergospherical said:
You made a small error in the rearrangement, but once corrected the last line is almost right; that's the general equation of a plane of normal ##\hat{n}##, but ##\hat{u} \cdot r## depends on ##t## so isn't constant in time (that's why the plane translates).
Whoops you're right, I wasn't very careful with my minus signs since I knew it was all going to be wrapped up into a constant anyway.

It looks like my equation is in the form of the scalar equation of a plane... is it really that simple? I have the equation of a plane so it is a plane wave?
 
  • Like
Likes DaveE and ergospherical
Blanchdog said:
It looks like my equation is in the form of the scalar equation of a plane... is it really that simple? I have the equation of a plane so it is a plane wave?
Yeah, pretty much 😄
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

Replies
8
Views
1K
Replies
8
Views
1K
Replies
2
Views
931
Replies
4
Views
4K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
2
Views
971