# Particle with gravity and electric field

1. Jun 25, 2010

### XanziBar

1. The problem statement, all variables and given/known data

A particle of mass m carries an electric charge Q and is subject to the combined action of gravity and a uniform horizontal electric field of strength E. It is projected in the vertical plane parallel to the field at a positive angle $$\theta$$ to the horizontal. Show that the horizontal distance it has traveled will be maximum if tan (2 $$\theta$$)=-mg/(EQ)

2. Relevant equations
Kinematics, F=QE etc.

3. The attempt at a solution

There is constant acceleration in y and constant acceleration in x and so I wrote down those equations and eliminated the time variable. I got the range of the projectile to be R=(2*V^2/g)* sin (theta)*(EQ*sin(theta)/(mg)+cos(theta))

So to get the max range I set all the cos/sin equal to one:

Rmax=(2*V^2/g)(EQ/(mg)+1)

So for my next step I tried setting the two equal to each other and solving for theta but there must be some kind of trig identity I am not seeing because I can not solve for theta or even figure out how to plug in tan(2 theta) to check the condition that I am supposed to prove. Help?

2. Jun 25, 2010

### collinsmark

Hello XanziBar,
You might want to show how you got to your range equation. I think there is a mistake somewhere (there are problems with your equation's dimensional units too, so there's definitely something not right) . I used the

$$s = v_0 t + \frac{1}{2}a t^2$$

kinematics equation for both x and y, and combined them to make a range equation. My range equation is different than yours however.
Ummm, :uhh: I do not follow what you mean by that approach.

Have you tried taking the derivative and setting it equal to zero? (That's usually how you maximize or minimize something.)
I'm still not quite sure of what your approach is there. But whatever the case, you should find this trigonometric identity very useful.

$$\mathrm{sin}(2 \theta) = 2 \mathrm{cos} \theta \mathrm{sin} \theta$$

Good luck!

3. Jun 25, 2010

### XanziBar

Alrighty well for the y direction I have

0=vsin(theta)t-.5*g*t^2

so v*sin(theta)=.5*g*t and that means that time in flight is t=2v*sin(theta)/g

for the x direction I have that delta x (range)=vcos(theta)*t+.5*Q*E*t^2/m where I went ahead and replaced the x acceleration with a=F/m=QE/m

So I plugged that expression for t into the x acceleration and got Range=cos(theta)*(2v*sin(theta)/g)+.5*Q*E*(2v*sin(theta)/g)^2/m

So after I simplify that is where I got my range equation, could you tell me where the units are wrong?
R=(2*V^2/g)* sin (theta)*(EQ*sin(theta)/(mg)+cos(theta)) because I think I simplified that right.

Anyway my approach that I was describing is that the biggest sine and cosine can be is unity so I figured that I could figure out what angle produces the maximum range.

Yeah I tried using 2sin(x)cos(x)=sin(2x) I didn't get anywhere but maybe I don't have the right intuition.

And yeah I did just try taking the derivative of the range w.r.t. theta...made a HUGE mess. Sorry if I'm not describing things well but I'm still really stuck...

4. Jun 25, 2010

### collinsmark

So far so good. Very nice.
I'm assuming that by 'v' you mean 'v0'. This term needs to be squared in the first instance of your above equation.
Okay. I see now. Your equation is correct. It's just hard to follow if your equations are completely in text form. Sorry for the confusion. For clarity let me retype your equation in LaTex:

Your equation (replacing v with v0 too):
$$R = \left(\frac{2 v_0^2}{g} \right) \mathrm{sin}\theta \left(\frac{EQ \mathrm{sin} \theta}{mg} + \mathrm{cos}\theta \right)$$

Your equation is correct, but let me give you some advice. Don't try to simplify things too much yet. Leave things un-factored for now. Let's go back to your unsimplified version of the equation.

$$R = \frac{2 v_0^2 \mathrm{cos}\theta \mathrm{sin}\theta}{g} + \frac{2 v_0^2 EQ \mathrm{sin}^2 \theta}{mg^2}$$

Now see if you can use that trigonometric identity somewhere,

$$\mathrm{sin}(2 \theta) = 2 \mathrm{cos} \theta \mathrm{sin} \theta$$

But that won't maximize anything. The sines and cosines are a necessary part of the equation, and it's not legal to just arbitrarily set them to 1 (well, at least not in this case, where there are a combination of both sines and cosines). By forcing the sines and cosines to be 1, you are essentially forcing θ to be 90o in parts of the equation, and 0o in other parts. But you don't necessarily want θ to be either 0o or 90o. You want it to be (generally) some other value -- whatever value that maximizes the range R.

Your goal is find the particular θ that maximizes the range R. To do this, find:

$$\frac{\partial R}{\partial \theta} = 0$$

(Take the derivative of your range equation with respect to θ, and set that equal to 0.)

Try it again now, in your unfactored equation. And keep in in your back pocket too, because you'll use it again later.
Try it again, after using the trig identity first.

5. Jun 25, 2010

### XanziBar

Yes! Thank you so much!

I don't know what I was thinking just setting things to 1 for no reason...haha it seems so ridiculous in retrospect.

It worked out just a couple of more lines after the derivative. Thanks so much. You were really helpful and clear!