What charge is carried by sphere c?

  • #1
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Homework Statement


http://imgur.com/a/wEUgn question #70 in the attached image.

Three charged spheres are at rest in a plane as shown in the figure. Spheres A and B are fixed, but sphere C is attached to the ceiling by a thread. The tension in the string is .240 N. Spheres A and B have charge Qa = 28 x 10^-9 and Qb = -28 x 10^-9.

What charge is carried by sphere C?

Homework Equations


Fe = (kQq)/r^2
f=ma


The Attempt at a Solution



This is a pretty hard question so please work with me guys.

I decided to start by listing all of the forces on sphere C

You have a tension force with N = .240, with the horizontal component being .240sin(30deg) left and vertical component being .240cos(30deg) up.

Since everything is at rest that means the horizontal and vertical tension forces would be canceled out right?

That would mean you have a force with magnitude .240sin(30deg) right and .240cos(30deg) down.

But that makes no sense. Because C has to be either + or -, which means it can only be either

1) .240sin(30deg) right and .240cos(30deg) up

or

2) .240sin(30deg) left and .240cos(30deg) down

I'm guessing I just don't understand the concept somewhere... can anyone please help me out?
 

Answers and Replies

  • #2
TSny
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You are understanding the concepts. Perhaps you left out a force and this is what is confusing you about the sign of charge C.
 
  • #3
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You are understanding the concepts. Perhaps you left out a force and this is what is confusing you about the sign of charge C.
I see! It must be the force of gravity then, correct?!
Ill keep on thinking about this question for a bit and come back if im having more doubts
 
  • #4
TSny
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I see! It must be the force of gravity then, correct?!
Ill keep on thinking about this question for a bit and come back if im having more doubts
Yes, good.
 
  • #5
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Yes, good.
Ok so I've been thinking about it but I still have some doubts.

So since i've figured out that it is a + force, there is only one force going in the right direction with magnitude .24sin(30deg)

so I set that = to the electrostatic force from B and my unknown C

[itex] .24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex]

Solve for C and that is my answer for the magnitude of its charge correct?
 
  • #6
ehild
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Ok so I've been thinking about it but I still have some doubts.

So since i've figured out that it is a + force, there is only one force going in the right direction with magnitude .24sin(30deg)

so I set that = to the electrostatic force from B and my unknown C

[itex] .24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex]

Solve for C and that is my answer for the magnitude of its charge correct?
The charge A does not exert any force on C? Why?
 
  • #7
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The charge A does not exert any force on C? Why?

Well it does, but it exerts a force in the up direction, along with the tension force, to counter the force due to gravity right?
 
  • #8
ehild
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Well it does, but it exerts a force in the up direction, along with the tension force, to counter the force due to gravity right?
No. Remember all charges are on a plane, so their forces of interaction are in that plane.
 
  • #9
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No. Remember all charges are on a horizontal plane, so their forces of interaction are horizontal.
Sorry I'm not getting what you mean.

Since the unknown charge is positive, and charge A is positive, it doesn't exert any force on the unknown charge right?

Because if it did the sphere would be going somewhere towards the top right, right?
 
  • #10
ehild
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The question is if that plane is horizontal or vertical?
Two positive charges do exert force on each other.
 
  • #11
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The question is if that plane is horizontal or vertical?

I would guess vertical because the string is attached to a ceiling, right? Maybe im confused which plane are you talking about.

No. Remember all charges are on a plane, so their forces of interaction are in that plane.

Are you trying to say that I need to find a way to find the electrostatic force in the up down axis?
 
  • #12
ehild
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I would guess vertical because the string is attached to a ceiling, right? Maybe im confused which plane are you talking about.



Are you trying to say that I need to find a way to find the electrostatic force in the up down axis?
I assumed a horizontal plane, with the ceiling above it.
If the plane is vertical, you are right, the force of charge A - gravity is equal to the vertical component of tension and the horizontal component of tension is equal to the force between C and B.
 
  • #13
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I assumed a horizontal plane, with the ceiling above it.
If the plane is vertical, you are right, the force of charge A + gravity is equal to the vertical component of tension and the horizontal component of tension is equal to the force between C and B.
the horizontal component of tension is equal to the force between C and B. wouldn't I be able to use equation

[itex] .24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex]

where k is a constant and c is the charge I'm trying to find, solve it for c, and that's my answer?
 
  • #14
ehild
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the horizontal component of tension is equal to the force between C and B. wouldn't I be able to use equation

[itex] .24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex]

where k is a constant and c is the charge I'm trying to find, solve it for c, and that's my answer?
Yes, but why is the minus sign there? You said charge C is positive.
 
  • #15
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Yes, but why is the minus sign there? You said charge C is positive.

Because charge B is -...

wait now I'm confused again... does that mean c is not +?

Assuming it didn't product any electrostatic field it would just hang down, but in this picture it is hanging to the right, so I took this to assume since that force is -, and there was some attraction that made it go right, that C is then a + force.

Either way its hanging down due to gravity, but since it is being attracted to the right and to the right is a - particle so c must be positive, then A has to be + too right? which means A is pushing away from C? so really its Fg and Fe = Force of tension on vertical plane

Sorry man I thought I figured this problem out but I just ended up confusing myself..
 
  • #16
ehild
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Better to show a picture.
upload_2017-7-15_9-47-38.png

QA is positive, it repulses charge C pushing it up with force Fa. The vertical component of tension is also up. Their sum has the same magnitude as the downward force of gravity: Tcos(30°) + Fa=G.
QB is negative, it attracts C to the right with the force Fb. The horizontal component of tension points to the left. Fb and the horizontal component of the tension are equal in magnitude.
 
  • #17
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Better to show a picture.
View attachment 207193
QA is positive, it repulses charge C pushing it up with force Fa. The vertical component of tension is also up. Their sum has the same magnitude as the downward force of gravity: Tcos(30°) + Fa=G.
QB is negative, it attracts C to the right with the force Fb. The horizontal component of tension points to the left. Fb and the horizontal component of the tension are equal in magnitude.

Okay this helps put me back where my thoughts were originally, ty.

Yes, but why is the minus sign there? You said charge C is positive.

Since charge C is positive, but b is negative, [itex] .24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex] the way I plugged it in follows that, I actually am suppose to mean:

[itex] -.24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex], right?
 
  • #18
ehild
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Since charge C is positive, but b is negative, [itex] .24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex] the way I plugged it in follows that, I actually am suppose to mean:

[itex] -.24sin(30deg) = (k(-28x10^-9)(c))/(.46)^2 [/itex], right?
It is right now.
 
  • #19
321
20
It is right now.

Thanks man. Took a while for me to get through what you were saying and I'm surprised I could have missed that.
 

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