Question about the motion of a charged particle

[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??

 

[Dale]

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?

 

[kuruman]

$\dots~$ 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?

 

[rehab]

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 ?

 

[rehab]

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

 

[kuruman]

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.

 

[rehab]

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?

 

[Dale]

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.

 

[kuruman]

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.

$\dots~$ if B was pointing in the Z direction and E in the y direction $\dots~$

 

[PeroK]

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

 

[kuruman]

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,

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 $~\dots$

 

[PeroK]

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!

 

[Steve4Physics]

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.

 

[kuruman]

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]

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!