Metal ball charged with electron orbiting

In summary, the electron has a charge of .02m and orbits a metal ball with a radius of .015m. The electric force keeps the electron in orbit. The formula for the centripetal acceleration is (9.11*10^-31*(12*10-6)^2 * .02^3)=9E9*1.6*10^-19*Q. The centripetal acceleration is provided by the tension in a connecting filament or string.
  • #1
conov3
34
0

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.
 
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  • #2
conov3 said:

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?
 
  • #3
ω angular rotation
angular velocity and then we would have a period of one rotation?
ω=2∏/T?
 
  • #4
Also, v=a^2/r
 
  • #5
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?
 
  • #6
? 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?
 
  • #7
Then find E=ma/q...

then q=Er/kR?
 
  • #8
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?
 
  • #9
Tension?
 
  • #10
conov3 said:
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?
 
Last edited:
  • #11
Centripetal force=mv^2/r
Centripetal acc.=ω^2*r
 
  • #12
conov3 said:
Centripetal force=mv^2/r
Centripetal acc.=ω^2*r

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

So, suppose you consider the formula for centripetal force, Fc = mω2r. What is supplying it? (What's holding the electron in orbit?)
 
  • #13
Are you meaning the charge of the metal ball? that's keeping it in place
 
  • #14
conov3 said:
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?
 
  • #15
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?
 
  • #16
conov3 said:
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:

[itex] F = k \frac{Q_1 Q_2}{r^2} [/itex]
 
  • #17
So from there I could say... mv^2/r=kqq/r^2
 
  • #18
so from there I could say mv^2/r=kQQ/r^2
 
  • #19
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
 
Last edited:
  • #20
conov3 said:
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:

[itex] m_e \omega^2 r = k \frac{Q q_e}{r^2} [/itex]

Where me is the electron mass and qe the magnitude of the charge on the electron.
 
  • #21
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!
 
  • #22
conov3 said:
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.
 
  • #23
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?
 
  • #24
conov3 said:
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 [itex] 2 \pi [/itex] radians per 12.0 μs, or 5.24 x 105 radians per second.
 
  • #25
okay I plugged it all in and got 1.39*10^-15C!
Thank you!
 
  • #26
conov3 said:
okay I plugged it all in and got 1.39*10^-15C!
Thank you!

Yup. That looks better :smile:
 
  • #27
Thanks a ton gneill! Better than a full class of learning haha :)
 

1. What is a metal ball charged with an electron orbiting?

A metal ball charged with an electron orbiting is a scenario in which a metal ball, typically a conducting material, has an excess or deficiency of electrons and is rotating around a center point due to the influence of an external electric field.

2. How does the electron orbiting around the metal ball affect its charge?

The electron orbiting around a metal ball can either increase or decrease its charge, depending on the direction of the electron's movement. If the electron is moving in the same direction as the electric field, it will increase the charge on the metal ball. If the electron is moving in the opposite direction of the electric field, it will decrease the charge on the metal ball.

3. What factors influence the speed of the electron orbiting around the metal ball?

The speed of the electron orbiting around the metal ball is influenced by the strength of the electric field, the distance between the electron and the metal ball, and the mass of the electron. The stronger the electric field, the closer the electron is to the metal ball, and the smaller the mass of the electron, the faster it will orbit.

4. Can the electron orbiting around the metal ball change direction?

Yes, the electron orbiting around the metal ball can change direction if the electric field changes direction. If the electric field changes direction, the electron's movement will also change, causing it to orbit in the opposite direction around the metal ball.

5. What is the significance of a metal ball charged with an electron orbiting?

The scenario of a metal ball charged with an electron orbiting is significant in understanding the behavior of charged particles in electric fields. It also has practical applications in devices such as particle accelerators and mass spectrometers, which use electric fields to manipulate and analyze charged particles.

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