Why Does a Dipole in E-Field Point in a Certain Direction?

[quietrain]

does anyone know why the forces point in such a direction?

basically, its a dipole (y-orient) in an E-field

but why do the forces on the +q and -q slant as such?

thanks

 

[Doc Al]

but why do the forces on the +q and -q slant as such?

The forces point in the direction of the field (or opposite to it). The field lines are curved and thus the forces are tangential to those field lines.

 

[jtbell]

That diagram is somewhat inaccurate. The angles between the force vectors and the line of the dipole should be smaller.

 

[Doc Al]

That diagram is somewhat inaccurate. The angles between the force vectors and the line of the dipole should be smaller.

Yes, they don't look quite as tangential as they should be. (Maybe that was the OP's point?)

 

[sophiecentaur]

If you think of the Magnetic Field around a short bar magnet (more familiar and often done in School) you would expect a similar field pattern. The field lines are pretty much circular for a short dipole. The Force is parallel with the field (which is how the field is defined, actually).

 

[quietrain]

oh but why is the force supposed to be tangential to the e-field at that point?

for example in this diagram, the force on the green + charge, shouldn't it be towards the - black charge?

why is the force tangential instead to the e-field?

 

[jtbell]

oh but why is the force supposed to be tangential to the e-field at that point?

By definition, the direction of the electric field is the direction of the electric force on a positively charged particle. In vector form,

$$\vec F_e = q \vec E$$

 

[quietrain]

ah i see thanks!

 

[shoestring]

If the dipole is free to rotate, it'll point in the opposite direction, with the positive end down and the negative end up. In that case the resulting force will be to the left.

 

[kame]

Is the dipole moving when the force go from right to left and back?

 

[quietrain]

If the dipole is free to rotate, it'll point in the opposite direction, with the positive end down and the negative end up. In that case the resulting force will be to the left.

i thought since the top wants to go bottom, the bottom wants to go top, the net force will be to the right?

as seen from the picture, the net force should be towards the right right?

 

[Doc Al]

as seen from the picture, the net force should be towards the right right?

Yes. With the dipole oriented as shown, the net force on it will point to the right.

 

[quietrain]

Yes. With the dipole oriented as shown, the net force on it will point to the right.

wow, so basically the dipole cannot align itself to the E-field but will shoot to the right?

so basically i can create a "gun" with such E-field and shoot charged dipoles out?

thats cool!

 

[jtbell]

If the vertical components of the two forces exactly balance each other, then the dipole will remain oriented as it is. However, this is an unstable equilibrium. If the dipole is even a tiny bit misaligned from the vertical, or if the field lines are not precisely circular arcs, then the dipole will start to rotate towards the opposite orientation, while moving to the right as previously noted.

Exercise: when the dipole has reversed its orientation, what is the direction of the net force now?

 

[shoestring]

So it could be used as some kind of gun shooting dipoles trough the opening to the left (unless there are fields outside the left opposing the effect). Or it could perhaps be used as a dipole trap. Is there anything like that in use already?

 

[quietrain]

If the vertical components of the two forces exactly balance each other, then the dipole will remain oriented as it is. However, this is an unstable equilibrium. If the dipole is even a tiny bit misaligned from the vertical, or if the field lines are not precisely circular arcs, then the dipole will start to rotate towards the opposite orientation, while moving to the right as previously noted.

Exercise: when the dipole has reversed its orientation, what is the direction of the net force now?

net force to the left? assuming the vertical components of the +ve(towards bottom) and the -ve(towards top) cancel out?

i just have to follow the E-field lines right? like you mentioned earlier

"By definition, the direction of the electric field is the direction of the electric force on a positively charged particle. "