# Homework Help: Charged particle moving under electric field

1. Apr 11, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
A charged particle having charge q and mass m is projected into a region of uniform electric field of strength E, with velocity v perpendicular to E. Throughout the motion apart from electric force particle also experiences a dissipative force of constant magnitude qE and directed opposite to its velocity. If |v|=6 m/s, then find its speed when it has turned through an angle of 90.

3. The attempt at a solution

The trajectory of the particle will be a curved path. Work done by dissipative force = ΔK.E.

$qEx=\dfrac{1}{2} m(v_f ^2 - v^2)$

But how do I find the path length traversed by the particle? It would be helpful if someone could guide me.

2. Apr 11, 2014

### TSny

Let the x-axis be in the direction of the initial velocity and the y-axis in the direction of the field.

Draw a diagram showing the two forces acting on the particle at some point of the trajectory. Draw the net force on the diagram. Consider the acute angle that the net force makes to the tangent of the trajectory and consider the angle that the net force makes to the y-direction. How do those two angles compare?

What does that tell you about dv/dt and dvy/dt, where v is the speed and vy the y-component of velocity?

3. Apr 11, 2014

### utkarshakash

I've got two equations

$qE(1- \sin \theta ) = m \frac{dv_y}{dt} \\ -qE \cos \theta = m \frac{dv_x}{dt}$

where θ is the angle made by the tangent with the +ve X-axis.

4. Apr 11, 2014

### TSny

OK, these equations look correct. But you can avoid having to deal with them if you continue along the lines that I suggested in my previous post.

From the vector diagram of the forces, can you see how the component of the net force along the tangent to the trajectory is related to the y-component of the net force?

5. Apr 12, 2014

### utkarshakash

Magnitude of net force = 2qEcosα

Magnitude of component of net force along tangent = Magnitude of y-component of net force = 2qEcos2α

where α is the angle made by the net force with vertical.

6. Apr 12, 2014

### TSny

Look at the two forces acting on the particle and the net force produced by these two forces. See the attached figure. Let $\small \alpha$ and $\small \beta$ be the angles between the net force and the tangential and y directions, respectively. From the fact that the two individual forces are equal in magnitude, what can you conclude about $\small \alpha$ and $\small \beta$?

#### Attached Files:

• ###### q forces.png
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152
7. Apr 12, 2014

### TSny

OK, good. This is the key point.

So what is the relation between dv/dt and dvy/dt?

8. Apr 12, 2014

### utkarshakash

α=β

$| \dfrac{dv}{dt} | = | \dfrac{dv_y}{dt} |$

But where does this lead me?

9. Apr 12, 2014

### TSny

Noting the directions of the forces, what does this relation look like after you remove the absolute values?

Try integrating the relation.

10. Apr 12, 2014

### utkarshakash

v=v_y. Does this mean that the velocity of particle will be 6 m/s?

11. Apr 12, 2014

### TSny

You didn't get quite get the relationship correct. First decide if dv/dt = dvy/dt or if dv/dt = -dvy/dt. (Consider the directions of the components of the net force along the tangential and y directions.)

If two functions have the same derivative, it doesn't necessarily mean that the functions are equal. But what can you say about them?

Last edited: Apr 12, 2014
12. Apr 12, 2014

### utkarshakash

I think it should be dv/dt=dvy/dt. From this I can only tell that their accelerations in two directions are equal.

13. Apr 12, 2014

### TSny

If you look at the figure in post #6, you can see that the net force (red vector) projected onto the tangent line is in the direction to slow the particle down. But the net force has a positive y-component. So, v is decreasing while vy is increasing. So, you need to write

dv/dt = -dvy/dt.

Or

d(v+vy)/dt = 0.

So, you have a quantity whose time derivative is zero. What can you conclude about that quantity?

14. Apr 13, 2014

### utkarshakash

It must be constant. Initially, v+v_y=v=6(as v_y=0). When the particle turns through 90°, the particle only possess vertical velocity. So, it must be 6, right?

15. Apr 13, 2014

### TSny

That looks good. So, v + vy = v0 at all times.

When the particle has only vertical velocity, how does v compare with vy?

16. Apr 13, 2014

### utkarshakash

I guess they become equal to each other. Is it so?

17. Apr 13, 2014

### TSny

Yes. So, what is the final speed (after a 90o turn)?

18. Apr 14, 2014

### utkarshakash

3m/s. But I'm still not satisfied why they would become equal to each other.

19. Apr 14, 2014

### Vibhor

I agree $\small \alpha = \small \beta$ .But I wonder how does that help us to deduce that $| \dfrac{dv}{dt} | = | \dfrac{dv_y}{dt} |$

I agree this is correct . I can derive it mathematically but not by looking at the diagram and from the fact that $\small \alpha = \small \beta$ .

$| \dfrac{dv}{dt} | ≠ qE$ .

Please help me understand how the equality of two angles lead us to $| \dfrac{dv}{dt} | = | \dfrac{dv_y}{dt} |$

20. Apr 14, 2014

### TSny

$\small v = \sqrt{v_x^2+v_y^2}$. What is $\small v_x$ after a 90o turn?

21. Apr 14, 2014

### utkarshakash

Nice. Thanks for having so much patience to guide me.

22. Apr 14, 2014

### TSny

Using the figure in post #6, consider the projection of the net force onto the tangent line of the trajectory. Compare that with the projection of the net force onto the vertical (y) direction.

What does Newton's second law then tell you about the tangential- and y-components of acceleration: at and ay?

How is tangential acceleration related to the rate of change of speed of the particle?

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Edit: Or you can use the figure attached here to see that the magnitude of the net force in the y-direction equals the magnitude of the net force along the tangent line.

#### Attached Files:

• ###### q forces 2.png
File size:
3.6 KB
Views:
100
Last edited: Apr 14, 2014
23. Apr 14, 2014

### utkarshakash

Your method is absolutely brilliant. May I know how did it occur to you at first glance? (just curious :tongue: )

24. Apr 14, 2014

### Vibhor

I am just trying to get to the root of my confusion .

Am I right if I say $$\dfrac{dv}{dt} ≠ qE$$ ?

25. Apr 14, 2014

### TSny

It was not at first glance at all. I did not see the simple approach based on the diagram until after I had blindly proceeded to solve the same differential equations that you found in post #3:

$qE(1- \sin \theta ) = m \frac{dv_y}{dt} \\ -qE \cos \theta = m \frac{dv_x}{dt}$

Dividing these, you get $\large \frac{dv_x}{dv_y} = -\frac{\cos\theta}{1-\sin\theta}$

Using $v_x = v\cos\theta$ and $v_y = v\sin\theta$ you can get a differential equation for $\frac{dv}{d\theta}$ which can be integrated to get the solution

$\large v = \frac{v_0}{1+\sin\theta}$.

This simple result made me step back and look for a more direct way to get the answer. Somewhere in working with the equations, I had noticed $\frac{dv}{dt} = -\frac{dv_y}{dt}$ and that this was the key to a simple solution. Then I realized that this relation could easily be deduced from the force diagram.

So, it was a very circuitous route. I wish I could say that I had the insight to see the simple solution straightaway, but I didn't.

Last edited: Apr 14, 2014