Question about the motion of a charged particle

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.
  • #1
rehab
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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??
 
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  • #2
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?
 
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  • #3
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?
 
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  • #4
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 ?
 
  • #5
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
 
  • #6
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.
 
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  • #7
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?
 
  • #8
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.
 
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  • #9
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~##
 
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  • #10
Doesn't the Lorentz force law say it all?$$\vec F = q(\vec E + \vec v \times \vec B)$$
 
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  • #11
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##
 
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  • #12
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!
 
  • #13
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.
 
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  • #14
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.
 
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  • #15
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!
 

1. What is the motion of a charged particle?

The motion of a charged particle refers to the movement of an electrically charged object in a given space. This motion is influenced by both electric and magnetic forces.

2. How does the charge of a particle affect its motion?

The charge of a particle determines the strength and direction of the electric and magnetic forces acting on it. A positive charge will be attracted to a negative charge and repelled by a positive charge, while a negative charge will be attracted to a positive charge and repelled by a negative charge.

3. What is the role of magnetic fields in the motion of a charged particle?

Magnetic fields can exert a force on charged particles, causing them to move in a circular or helical path. This is known as the Lorentz force and is responsible for the circular motion of charged particles in a magnetic field.

4. How does the speed of a charged particle affect its motion?

The speed of a charged particle can affect its motion in several ways. A higher speed can increase the strength of the magnetic force acting on the particle, causing it to move in a wider arc. Additionally, the speed of a charged particle can determine whether it will follow a straight or curved path in a magnetic field.

5. What factors can influence the motion of a charged particle?

The motion of a charged particle can be influenced by several factors, including the strength and direction of electric and magnetic fields, the charge and mass of the particle, and its initial velocity. Other factors such as collisions with other particles or interactions with other forces can also affect the motion of a charged particle.

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