What Speed Must an Electron Achieve to Orbit a Charged Glass Sphere?

Click For Summary
SUMMARY

The speed required for an electron to orbit a charged glass sphere with a diameter of 2.70 mm and a charge of +1.10 nC at a distance of 1.50 mm above the surface is calculated using the electrostatic force and uniform circular motion equations. The correct radius for calculations includes both the radius of the sphere and the separation distance, leading to a final speed of approximately 2.469 x 107 m/s. This value is derived from the formula involving the mass of the electron (9.10938 x 10-31 kg), Coulomb's constant (8.99 x 109 N m2/C2), and the charges involved. The discussion emphasizes the importance of accurately determining the effective radius in electrostatic calculations.

PREREQUISITES
  • Understanding of electrostatic force calculations using Coulomb's law
  • Familiarity with uniform circular motion equations
  • Knowledge of electric field concepts and their relationship to force
  • Basic grasp of physics constants such as the charge of an electron and Coulomb's constant
NEXT STEPS
  • Study the derivation of Coulomb's law and its applications in electrostatics
  • Learn about electric fields generated by charged objects, specifically uniformly charged spheres
  • Explore the relationship between force, mass, and acceleration in circular motion
  • Investigate the implications of varying distances in electrostatic interactions
USEFUL FOR

Students and enthusiasts of physics, particularly those studying electrostatics and circular motion, as well as educators seeking to clarify concepts related to charged particles and their dynamics.

tomizzo
Messages
113
Reaction score
2

Homework Statement



A 2.70-mm-diameter glass sphere has a charge of + 1.10 nC.

What speed does an electron need to orbit the sphere 1.50mm above the surface?

Homework Equations



a = v^2/r -> force = m*v^2/r

electrostatic force = K*Q1*Q2/distance^2

therefore:

m*v^2/r=K*Q1*Q2/distance^2

The Attempt at a Solution



(9.10938291 × 10-31 kg)(v^2)/((2.7*10^-3)/2) = 8.99*10^9*(1.60*10^-19)*(1.1*10^-9)/(1.50*10^3)^2

I get 32282518 m/s which is the incorrect answer. However, I just noticed something. The radius I'm using is half of the sphere, should I be adding the separation distance to the radius as well?
 
Physics news on Phys.org
yes, the circle's radius is bigger than the glass radius.
 
lightgrav said:
yes, the circle's radius is bigger than the glass radius.

I got an answer of 4.69*10^7 m/s and it still says it's incorrect.
 
I didn't get that speed. why don't you cancel one of the radius variables, and try the calculation again.
 
Here are my values:

Mass of electron = 9.10938*10^-31 kg
v = ?
radius = half of glass sphere plus separation distance = (2.7/2)*10^-3+1.5*10^-3
K = 8.99*10^9
Q1 = 1.6*10^-19 (charge of electron in coulombs)
Q2 = 1.1*10^-9
distance = 1.5*10^-3

Now I'm realizing the distance should probably be (2.7/2)*10^-3+1.5*10^-3 also...
 
With the change, I'm getting an answer of 2.469*10^7 m/s. Does this look familiar?
 
I think it rounds up to 24.7 Mm/s ... the E-field has spread more at that distance, so is weaker at the electron.
 
  • Like
Likes   Reactions: 1 person
Well, this is obviously way too late, but for all of you physics nerds out there here it is.

Uniformly Charged Sphere Equation:

E=Q/(e0*4*r^2*pi)

E=electric field
Q=charge of sphere 1.1*10^-9 (in this case)
e0=8.85*10^-12
r=1.5*10^-3 (in this case)
pi = pi ;)

Take that E and plug it into the following Electric Force Equation:

E=F/q

E= what you solved for previously
F=Force that you want to derive
q=1.6*10^-19 (constant for charge of electron)

Take that F and plug it into this standard Force Equation:

F=ma

F=what you solved for previously
m=9.11*10^-31
a=what you want to solve forLast, but not least plug the a you solved for into the following Uniform Circular Motion Equation:

a=v^2/r

a=what you just solved for
v=what you want to solve for
r=1.5*10^-3 (in this case)That's it ladies and gents. Physics = MAGIC
 

Similar threads

Replies
2
Views
9K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
3K
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K