Metal ball charged with electron orbiting

[conov3]

Homework Statement

An electron orbits a 3.0 cm diameter metal ball 5.0 mm above the surface. The orbital period of the electron is l2.0 micro seconds. What is the charge on the metal ball?

r=1.5cm or .015m
R=.02m (.015+.005)
Orbit=12*10^-6s

Homework Equations

E=(kq/r)R which gives electric field
k=9*10^9
Maybe F=qE?
Is this the only equation?

The Attempt at a Solution

(9*10^9*1.6*10^-19/.015)*.02 which gives E. not sure what to do next!

I fell like the time is involved with the problem.. but do not know where it takes part and cannot find another formula in my book.

 

[gneill]

Homework Statement

An electron orbits a 3.0 cm diameter metal ball 5.0 mm above the surface. The orbital period of the electron is l2.0 micro seconds. What is the charge on the metal ball?

r=1.5cm or .015m
R=.02m (.015+.005)
Orbit=12*10^-6s

Homework Equations

E=(kq/r)R which gives electric field
k=9*10^9
Maybe F=qE?
Is this the only equation?

The Attempt at a Solution

(9*10^9*1.6*10^-19/.015)*.02 which gives E. not sure what to do next!

I fell like the time is involved with the problem.. but do not know where it takes part and cannot find another formula in my book.

You have an orbiting body; it's traveling in a circular orbit. What forces act on a body in circular motion?

 

[conov3]

ω angular rotation
angular velocity and then we would have a period of one rotation?
ω=2∏/T?

 

[conov3]

Also, v=a^2/r

 

[gneill]

Well, you're on the right track. It looks like you know how to find the angular velocity. What force is required to maintain circular motion? What is providing that force?

 

[conov3]

? kind of confused with that question
Wouldnt that be angular acceleration?
so
a=v^2/r
I actually switched my a and v on the last post. I meant to give angular acceleration.
So from there, if I find my acceleration.. I can say F=ma? then ma=qE?

 

[conov3]

Then find E=ma/q...

then q=Er/kR?

 

[gneill]

Yes, you can balance forces or accelerations -- they are proportionally related via the mass of the orbiting object.

No, it's not angular acceleration that you want. What acceleration is involved in circular motion?

 

[conov3]

Tension?

 

[gneill]

Tension?

Tension is a force in a connecting filament or string. No, you're looking for either centripetal acceleration or centripetal force (When an object is whirled around on the end of a string, the centripetal force is provided by tension in the string).

What are the formulas for centripetal force and centripetal acceleration?

 

[conov3]

Centripetal force=mv^2/r
Centripetal acc.=ω^2*r

 

[gneill]

Centripetal force=mv^2/r
Centripetal acc.=ω^2*r

Yup. You might want to note that centripetal acceleration is also a_c = v²/r . It gets made into centripetal force by multiplying by the mass: f = ma. Similarly, centripetal force can be had as: F_c = mω²r.

So, suppose you consider the formula for centripetal force, F_c = mω²r. What is supplying it? (What's holding the electron in orbit?)

 

[conov3]

Are you meaning the charge of the metal ball? that's keeping it in place

 

[gneill]

Are you meaning the charge of the metal ball? that's keeping it in place

Yes, what force does the charged ball exert on the orbiting electron? What's the formula?

 

[conov3]

E=(kq/r)R? I am not sure this is where I struggled in our first phys semester. I know that the centripetal acceleration of the electron would point towards the ball..
but E does not equal F so that's not the right equation right?

 

[gneill]

E=(kq/r)R? I am not sure this is where I struggled in our first phys semester. I know that the centripetal acceleration of the electron would point towards the ball..
but E does not equal F so that's not the right equation right?

Your text or notes should have the equation for the electric force:

F = k \frac{Q_1 Q_2}{r^2}

 

[conov3]

So from there I could say... mv^2/r=kqq/r^2

 

[conov3]

so from there I could say mv^2/r=kQQ/r^2

 

[conov3]

or do i use ω instead of v^2/r and it equals 12*10^-6

If this is correct, I got a final answer or q=3.644*10^-35C

 

[gneill]

so from there I could say mv^2/r=kQQ/r^2

Keep in mind that there are two charges involved -- the charge of the ball and the charge on the electron. So write 'Qq' rather than 'QQ'.

You can use either formula for the centripetal acceleration. One requires you to find the angular velocity, the other the tangential speed of the electron. You've already shown that you can find the angular velocity quite easily... So your equation equating the centripetal force to the electrical force becomes:

m_e \omega^2 r = k \frac{Q q_e}{r^2}

Where m_e is the electron mass and q_e the magnitude of the charge on the electron.

 

[conov3]

Yes, so I use... (9.11*10^-31*(12*10-6)^2 * .02^3)=9E9*1.6*10^-19*Q
I didn't realize the squared radius on the right side at first so I need to recalculate this answer.
so Q=7.288*10^-37C
Thank you so much for all the help!

 

[gneill]

Yes, so I use... (9.11*10^-31*(12*10-6)^2 * .02^3)=9E9*1.6*10^-19*Q
I didn't realize the squared radius on the right side at first so I need to recalculate this answer.
so Q=7.288*10^-37C
Thank you so much for all the help!

How did you determine your values for ω and r? They don't look right to me.

 

[conov3]

I thought ω=12*10^-6 because an orbital period is 12microseconds.
For r I used the center point of the metal ball. The radius of the metal ball would be .015m and out to the electron would be an extra .005m? Do I also need to add the radius of an electron to .02?

 

[gneill]

I thought ω=12*10^-6 because an orbital period is 12microseconds.
For r I used the center point of the metal ball. The radius of the metal ball would be .015m and out to the electron would be an extra .005m? Do I also need to add the radius of an electron to .02?

Okay, your orbital radius is fine at 0.02m (my mistake, I thought I saw the problem state 3.0 cm as the radius of the ball rather than the diameter). The angular velocity though, should reflect 2 \pi radians per 12.0 μs, or 5.24 x 10⁵ radians per second.

 

[conov3]

okay I plugged it all in and got 1.39*10^-15C!
Thank you!

 

[gneill]

okay I plugged it all in and got 1.39*10^-15C!
Thank you!

Yup. That looks better

 

[conov3]

Thanks a ton gneill! Better than a full class of learning haha :)