What is the magnitude of charge on each sphere?

In summary, the problem involves two small metallic spheres with the same electric charge suspended by light strings at an angle of 5 degrees with the vertical. The distance between the spheres is determined to be 0.0523m and the weight of each sphere is found to be 0.00196N. Using Coulomb's law, the charge on each sphere is calculated to be 7.2 nC.
  • #1
AznBoi
471
0

Homework Statement


Two small metallic spheres, each of mass 0.2g are suspendedas pednulums by light strings from a common point. The spheres are given the same eletric charge, and it is found that they come to equilibrim when each stringis at an angle of 5degrees with the vertical. If each string is 30 cm long, what is the magnitude of the charge on each sphere?


Homework Equations


F=kq1q2/r^2


The Attempt at a Solution



I figured I would need the distance between the spheres so I used x=.3m sin5 deg and multiplied x by 2 to get the distance (r) between them.

Do I add two force equations and put them equal to zero? What other equations do I need? If I try solving for q that way everything cancels out. :confused:
 
Physics news on Phys.org
  • #2
Ok, so you have the distance, and you want the charge, so you need the force repelling these spheres.
Code:
 /\           y
O  O          |_x
Just consider one of these balls. You know that the electric force is keeping them apart, but there must be a force stopping them from flying apart in the x direction. Gravity points downwards (y direction), so it's not that.

The only other thing it could be is the string. The string's force (tension) can be broken up into components. One of these must be equal and opposite to gravity, and the other must be equal and opposite to the EM force, otherwise the spheres would not be stationary.

You know the gravitational force, so you know one of the components of the string. You can then find the other component of the string, which should give you what you need.

Hope this helps.
 
  • #3
Hmm, I kind of get what you are saying. So I need to make the x component of the tension force equal to the outward eletric force?

I think I will be able to find tension because that's what the mass is for! I'll try it out, thanks! :smile:
 
  • #4
I still didn't manage to get the correct answer.

Here's my work:

I found the (r) distance between the two metal spheres first.
[tex]sin5deg= \frac {5}{.03m}[/tex]
[tex]2x=r=.0523m[/tex]

I drew a FBD with <--- as F_e , ----> as T_x , Up vector as T_y and Down vector as the weight.

[tex] T_y=W=mg=(2*10^-4kg)(9.8)= 0.00196 N [/tex]

Used sine/cosine to find T_x and got: T_x=1.7147*10^-4 N

Tx-F_e=0 so Tx=F_e

Here is where I think I've messed up but I'm not sure:

[tex] T_x=k_e\frac{2q}{r^2}[/tex]

[tex]q=\frac{T_x*r^2}{2K_e}[/tex]

and I solved for q and I got 2.6 x 10^-17

The correct answer is 7.2 nC. What am I doing wrong? Thanks!
 
  • #5
It looks pretty at first glance, up until you get to 2Q, think that should be
q^2
 
  • #6
denverdoc said:
It looks pretty at first glance, up until you get to 2Q, think that should be
q^2

ooh. but I thought 2q and q^2 are the same? I'll try it anyways, thanks! :smile:
 
  • #7
AznBoi said:
ooh. but I thought 2q and q^2 are the same? I'll try it anyways, thanks! :smile:

Er, no.

q+q = 2q
q*q = q2
 
  • #8
Woww! It works! Can you explain to me why I can't use 2q instead of q^2.

Edit: oh ok. xP I'll try to not make a mistake like that anymore.Thanks you guys! =D
 
  • #9
AznBoi said:
Woww! It works! Can you explain to me why I can't use 2q instead of q^2.

Because coulomb's law needs the product of the two charges and not the sum.
[tex]k \frac{q1 \cdot q1}{r^2} \not= k \frac{q1+q1}{r^2}[/tex]
[tex]q \cdot q \not = q+q
[/tex]

When we have several of the same numbers, x, being multiplied together. It is more efficient to do xn, where n is the number of x's we have.
If we have several of the same numbers, x, being added together. It is more efficient to do n*x, where n is the number of x's being added.
 
  • #10
ranger said:
Because coulomb's law needs the product of the two charges and not the sum.
[tex]k \frac{q1 \cdot q1}{r^2} \not= k \frac{q1+q1}{r^2}[/tex]
[tex]q \cdot q \not = q+q
[/tex]

When we have several of the same numbers, x, being multiplied together. It is more efficient to do xn, where n is the number of x's we have.
If we have several of the same numbers, x, being added together. It is more efficient to do n*x, where n is the number of x's being added.

Thanks for the awsome explanation! :smile: I completely get it now.
 

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of electrostatics that describes the force between two electrically charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. What is the problem with Coulomb's Law?

The main problem with Coulomb's Law is that it assumes that the charges are stationary, which is not always the case in real-world situations. In reality, charges can move and change position, which can affect the accuracy of the calculated force.

3. How does the problem with Coulomb's Law affect its applications?

The problem with Coulomb's Law can lead to inaccurate predictions and calculations in situations where the charges are not stationary. This can be problematic in fields such as electronics and telecommunications, where precise calculations are necessary.

4. What are some solutions to the problem with Coulomb's Law?

One solution is to use the concept of electric fields, which takes into account the movement of charges. Another solution is to use a modified form of Coulomb's Law, known as the Lorentz Force Law, which incorporates the effects of both electric and magnetic fields on moving charges.

5. Is Coulomb's Law still a valid law despite its problem?

Yes, Coulomb's Law is still a valid law in situations where the charges are stationary. It is a fundamental law of electrostatics and has been extensively tested and verified through experiments. However, in situations where the charges are moving, other laws or modifications of Coulomb's Law should be used for more accurate results.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
927
  • Introductory Physics Homework Help
Replies
6
Views
768
Replies
17
Views
881
  • Introductory Physics Homework Help
Replies
21
Views
608
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
701
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
985
  • Introductory Physics Homework Help
Replies
6
Views
609
  • Introductory Physics Homework Help
Replies
5
Views
639
Back
Top