Question about the motion of a charged particle

• I
• rehab
In summary: It just is. It could be either positive or negative.I believe the correct statement is: If ##q>0##, then ##\vec F## is in the same direction as ##\vec E##; if ##q<0##, then ##\vec F## is in the opposite direction as ##\vec E##. This is because the force is in the **opposite** direction as the electric field if the charge is negative.

rehab

Hi, I have a question about the motion of a charged particle in crossed E and B fields. if B was pointing in the Z direction and E in the y direction then the component of the motion in the Z plane = 0. The only reason for this to happen is that the electric force due to the E field depends on the dot product of the E field and the velocity but the electric force = qE which is independent of the velocity, so why is the motion in the Z plane equal zero??

rehab said:
if B was pointing in the Z direction and E in the y direction then the component of the motion in the Z plane = 0
This is not true. The charged particle can have a component of velocity in any direction, completely regardless of the fields. Why do you think that it could not?

vanhees71
rehab said:
##\dots~## so why is the motion in the Z plane equal zero??
The acceleration in the Z direction is zero. That does not mean that the "motion" in the Z direction is zero. Are you confusing acceleration and velocity?

vanhees71 and Dale
Dale said:
This is not true. The charged particle can have a component of velocity in any direction, completely regardless of the fields. Why do you think that it could not?
here I'm not talking about the component of velocity , I'm asking about why the total motion=force in the z direction is zero ?

kuruman said:
The acceleration in the Z direction is zero. That does not mean that the "motion" in the Z direction is zero. Are you confusing acceleration and velocity?
Yeah I know that but I meant the total motion in Z direction due to the fields , regardless of the velocity

rehab said:
here I'm not talking about the component of velocity , I'm asking about why the total motion=force in the z direction is zero ?
If the magnetic field is in the z-direction, the magnetic force has zero component along the z-direction.
If the electric field is in the y-direction, the electric force has zero component along the z-direction.
The sum of two zeroes is identically zero.

vanhees71 and Orodruin
kuruman said:
If the magnetic field is in the z-direction, the magnetic force has zero component along the z-direction.
If the electric field is in the y-direction, the electric force has zero component along the z-direction.
The sum of two zeroes is identically zero.
my question is why the electric force has a zero component along the z-direction if it does not depend on the dot product of the E field and the velocity in Z?

rehab said:
here I'm not talking about the component of velocity , I'm asking about why the total motion=force in the z direction is zero ?
This is very confusing. Motion does not equal force. Motion is velocity. They are two very different concepts.

If ##\vec B = (0,0,B_z)## and ##\vec E = (0,E_y,0)## then the force on a charge ##q## moving with velocity ##\vec v=(v_x,v_y,v_z)## is given by $$\vec F = q \vec E + q \vec v \times \vec B = q (B_z v_y, E_y - B_z v_x, 0)$$ The force from the E field is in the direction of the E field and the force from the B field is in the direction perpendicular to both the B field and the velocity. So neither field produces a force in the z direction in this setup.

More generally, if B and E are perpendicular to each other then you will never get a force in the direction of the B field. This is because B produces forces perpendicular to B and E produces forces parallel to E. So if they are perpendicular to each other then no component of the total force will be in the direction of B.

Last edited:
PeroK, topsquark and vanhees71
rehab said:
my question is why the electric force has a zero component along the z-direction if it does not depend on the dot product of the E field and the velocity in Z?
Because the E field is along the y-direction. You said so yourself. It cannot have a z-component.
rehab said:
##\dots~## if B was pointing in the Z direction and E in the y direction ##\dots~##

vanhees71
Doesn't the Lorentz force law say it all?$$\vec F = q(\vec E + \vec v \times \vec B)$$

vanhees71, Dale and topsquark
PeroK said:
Doesn't the Lorentz force law say it all?$$\vec F = q(\vec E + \vec v \times \vec B)$$
Yes, it does. However, it seems that OP is confused thinking that the electric force is and is not velocity-dependent,
rehab said:
The only reason for this to happen is that the electric force due to the E field depends on the dot product of the E field and the velocity but the electric force = qE which is independent of the velocity ##~\dots##

vanhees71 and PeroK
rehab said:
my question is why the electric force has a zero component along the z-direction if it does not depend on the dot product of the E field and the velocity in Z?
I'm sorry to say that I don't understand anything you've written in any of your posts. Sorry!

rehab said:
my question is why the electric force has a zero component along the z-direction if it does not depend on the dot product of the E field and the velocity in Z?
@rehab, you may be confused due to an incorrect concept of some sort.

When a charge (##q##) is in an electric field (##\vec E##), the charge experiences an electric force (##\vec {F_E})##.

##\vec {F_E} = q\vec E## (note that ##q## is a scalar)

That means ##\vec {F_E}## is always in the same direction as ##\vec E##.

[Edit/correction: as pointed out by kuruman in Post #14, if ##q## is negative then ##\vec {F_E}## is in the opposite direction to ##\vec E##.]

The direction of the charge's velocity (##\vec v##) is completely unrelated to this.

Maybe you are confusing this with power (##P = \vec F \cdot \vec v##) which is a different issue.

Last edited:
rehab
Steve4Physics said:
That means ##\vec {F_E}## is always in the same direction as ##\vec E##.
You probably meant to say that ##\vec {F_E}## is always in the same direction as ##q\vec E##. The charge is not known to be positive.

Steve4Physics and hutchphd
kuruman said:
You probably meant to say that ##\vec {F_E}## is always in the same direction as ##q\vec E##. The charge is not known to be positive.
Yes indeed!