Homework help:A small plastic sphere with a mass of 2.60 X 10-15 kg

Click For Summary

Homework Help Overview

The problem involves a small plastic sphere suspended between two charged plates, with a focus on determining the type and magnitude of charge on the sphere, as well as the number of excess or deficit electrons. The context includes concepts from electrostatics and forces acting on charged objects.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the reasoning behind the charge type on the sphere, questioning how the forces balance to achieve suspension. There are attempts to calculate the charge and the number of electrons based on given equations and parameters.

Discussion Status

Some participants have provided feedback on calculations, noting potential errors in the exponent of the charge value. There is an ongoing exploration of the implications of these calculations on the charge's nature and the number of electrons involved.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can share or the methods they can use. There is a focus on ensuring the reasoning aligns with the physical principles involved.

Jaimie
Messages
35
Reaction score
0

Homework Statement



"A small plastic sphere (with a mass of 2.60 X 10-15 kg) is suspended between
two plates, with a potential difference of 265.4 V across the plates. The separation of the plates is
0.500 cm. The upper plate is negative and the lower one is positive.
a)What type of charge is on the sphere?Explain your reasoning.
b)What is the magnitude of the charge on the sphere?
c)How many excess or deficit electrons does the sphere have?"

Homework Equations



q=mgd/∆V
q=Ne

The Attempt at a Solution



a) The charge on the sphere is positive. In order for the sphere to suspend, there must be a net force of zero on the object, where the magnitude of the upwards electrical force is equal to the downwards gravitational force. As the electrical force is pointed upwards (acting in the direction of the electrical field), it must be attracted to the negatively-charged plate. If the sphere’s charge were negative, the electrical force would be repelling from the negative plate and pointed downwards, where the net force acting on the object would no longer be zero.

b) Fe= εq; ε= ∆V/d
Fe= Fg
(∆V/d)(q)= mg
∴q=mgd/∆V=2.60 x 10^-15 kg(9.8N/kg)(0.500cm)(1m/100cm)/265.4V=4.8 x 10^-4 C

c) q=Ne
q/e=N
∴N= q/e= 4.8 x 10^-4 C /(1.6 x 10^-19 C) =3.0 x 10^15 electrons
∴Deficit of 3.0 x 10^15 electrons, as charge is positive.

Please advise if this is correct?
 
Physics news on Phys.org
Jaimie said:
(∆V/d)(q)= mg
∴q=mgd/∆V=2.60 x 10^-15 kg(9.8N/kg)(0.500cm)(1m/100cm)/265.4V=4.8 x 10^-4 C

Your method looks good. But check to see if you really get a power of -4 in the answer for the charge.
 
Oh...calculation error...this should be 4.8 x 10^-19 C. TSny, thank you for letting me know.
 
So then for c)
q=Ne
q/e=N
∴N= q/e= 4.8 x 10^-19 C /(1.6 x 10^-19 C) =3.0 electrons
∴Deficit of 3.0 electrons, as charge is positive.
 
That looks good.
 
Thank you for your help!
 

Similar threads

Replies
3
Views
8K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K