1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Time lag before application of force

  1. Aug 30, 2016 #1
    Hi everyone,
    I present a situation here: Say there are two points A and B in space separated by a distance d. I create two charged particles (of any charge) at points A and B at time t=0. After how long will they experience force due to one another?

    My answer: We know that electric fields travel at c (the speed of light= 3 x 108 m/s). So, the field from the charge at A will spread out in a spherical way at time t=0. Similarly, the field from B will spread out spherically at the same time t=0. These two field will meet each other at the midpoint on the line joining A and B at a time=d/(2c). We know that interaction of fields leads to force. Field is the cause and force is the effect. Hence, according to me, the answer is t=d/(2c).

    Am I correct? If not, where am I going wrong?

  2. jcsd
  3. Aug 30, 2016 #2


    User Avatar
    Gold Member

    I speculate you are off by a factor of 2 if your thought-experiment detector is actually one of the particles. A particle can't 'know' about a value change at some specific point in a field faster than c, I think.
  4. Aug 30, 2016 #3
    That is correct. In fact there are a couple of questions here. You will need to specify clearly how you can "create" two charges at two different places, simultaneously.
  5. Aug 30, 2016 #4


    Staff: Mentor

    This is impossible, both for classical EM and also for quantum EM.
  6. Aug 31, 2016 #5
    I meant to say that I am willing to find after how much time the effect of the field created by both the charges at time t=0 will be felt by the charges, i.e. after how long will the charges feel force acting on them due to the field created by them at time t=0. Assuming that the charges were present from before can be done safely.

    This can also be put in a different way. Say, the field created by the charges were being shielded initially and at time t=0, the shield is removed. Now, at what time will force due to the other charge start acting?

    And, according to me it should be t=d/(2c). This is because force acts due to interaction of fields (and not due to interaction of field and a charge) and the fields interact for the first time at t=d/(2c).

  7. Aug 31, 2016 #6


    User Avatar
    Science Advisor

    Would violate SR.
  8. Aug 31, 2016 #7
    I think you are confusing some concepts here, the forces acting as a result of an interaction of fields is a QFT thing I think ( admittedly I don't know much about QFT so someone might want to correct me) and the fields they talk about are not the same as the force field you are imagining, for the situation you are describing ( disregarding the impossibility of it) you need the field to get a non zero value at the point where the charge is, so t =d/c
  9. Aug 31, 2016 #8


    Staff: Mentor

    This is essentially possible. Shielding is not magic, it is just opposite charges which cancel out the field of the first charge. So the simplest form of shielding would be a co-located opposite charge which moves away suddenly, forming a dipole at t=0.

    With that, your question would be about the timing of the interaction between the two dipoles. If the two dipoles are a distance d apart, then each will feel a force from the other at t=d/c.
    Last edited: Aug 31, 2016
  10. Aug 31, 2016 #9
    My answer would be that it would be t=d/c, it is not the interaction between the fields which moves the particles, it would be the particles interacting with the field, so you'd have to wait until the particle senses a change in the field.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted