# I Time lag before application of force

1. Aug 30, 2016

### americast

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?

Gramercy...

2. Aug 30, 2016

### Grinkle

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.

3. Aug 30, 2016

### Chandra Prayaga

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.

4. Aug 30, 2016

### Staff: Mentor

This is impossible, both for classical EM and also for quantum EM.

5. Aug 31, 2016

### americast

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).

Gramercy...

6. Aug 31, 2016

### A.T.

Would violate SR.

7. Aug 31, 2016

### nashed

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

8. Aug 31, 2016

### 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
9. Aug 31, 2016

### DarkBabylon

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.