Rate of decreasing static field - is it measurable?

  • #1
107
0

Main Question or Discussion Point

If a static charge is placed on an object, and that object begins to move through space, a magnetic field will form around the object (relative to a stationary observer).

At the same time, the static field around the object will begin to reduce. The faster the object moves, the stronger the magnetic field will become and the weaker the static force.

How do we measure that rate of the changing static force? Is this rate completely theoretical (I believe it's something like P=E.H.sineo), or has anyone managed to actually measure the static force at high speeds to 'actually' measure this change? Where could I find those results?

Thanks!
:)
 
Last edited:

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
2,894
1,232
If a static charge is placed on an object, and that object begins to move through space, a magnetic field will form around the object (relative to a stationary observer).

At the same time, the static field around the object will begin to reduce. The faster the object moves, the stronger the magnetic field will become and the weaker the static force.

How do we measure that rate of the changing static force? Is this rate completely theoretical (I believe it's something like P=E.H.sineo), or has anyone managed to actually measure the static force at high speeds to 'actually' measure this change? Where could I find those results?

Thanks!
:)
Actually, I don't that's quite right. The (static) electric field is not a function of velocity.

[tex] \vec E = \frac{1}{4 \pi \epsilon _0}\frac{q}{r^2} \hat r [/tex]

[tex] \vec F_E = q\vec E [/tex]

The magnetic field is, on the other hand.

[tex] \vec B = \frac{\mu _0}{4 \pi}\frac{q (\vec v \times \hat r)}{r^2} [/tex]

[tex] F_M = q(\vec v \times \vec B) [/tex]

Thus the total force is,

[tex] \vec F = \vec F_E + \vec F_M [/tex]

[tex] = q \vec E + q(\vec v \times \vec B) [/tex]

If you calculate the relative strengths between the electric and magnetic force (as a function of v) you get an interesting result. Note that Maxwell's equations tell us that the speed of light in a vacuum is [tex] c = \frac{1}{\sqrt{\epsilon _0 \mu _0}} [/tex].
 
  • #3
107
0
Were not E and H multiplied by each other supposed to equal a constant?
 
  • #4
collinsmark
Homework Helper
Gold Member
2,894
1,232
Were not E and H multiplied by each other supposed to equal a constant?
Something like

[tex] \vec E \cdot \vec H = \vec E' \cdot \vec H' [/tex]

Yeah, but I think that's more to do with the geometry changing when you go from one from one inertial frame to another inertial frame (at a different velocity) due to Lorenz transformations.

Yeah, I think I see what you mean now. But I'll let others comment from here.
 

Related Threads on Rate of decreasing static field - is it measurable?

Replies
4
Views
6K
Replies
20
Views
2K
Replies
6
Views
345
Replies
2
Views
1K
Replies
0
Views
1K
  • Last Post
Replies
4
Views
32K
Replies
5
Views
6K
Replies
1
Views
593
Replies
12
Views
3K
Top