Electric Charge and Electric Field

  • Thread starter Mdhiggenz
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  • #1
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Homework Statement


1zgx8vs.jpg


Two tiny spheres of mass = 6.60mg carry charges of equal magnitude, 72.0nC , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530m . When a horizontal uniform electric field that is directed to the left is turned on, the spheres hang at rest with the angle theta between the strings equal to 50° in the following figure.

What is the magnitude E of the field?



Homework Equations





The Attempt at a Solution



I know that q is moving towards the left towards the electric field.

I know that the tension in the x direction is Tx=Hcos∅

and in the y direction is Ty=hsin∅

I know the sum of the forces in the x direction ƩFx= hcos∅ + q

While the sum of the forces in the y direction is ƩFy=hsin∅-mg

Here is where i'm confused I have no idea how to find q. Would q just be 72?

In which case I get the magnitude of √ ƩFx^2 + ƩFy^2 = F0

then simply divde Fo/q to get E ?

i'm pretty confused some guidance would be awesome.

Thanks

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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92
We aren't asked about tension, so its better we won't deal with that in this case.
Start by finding the net force in x-direction on the left sphere. Then you can use the trig ratio
tanθ=Fx/Fy.
 
  • #3
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That's where I am confused the net force of the x direction involves q but I have no idea how they want me to get that q. Also how would the trig ratio tan x/y help me in this case. They are asking for the field.
 
  • #4
327
1
Ohh I see, I'm guessing we can use the trig ratio to find the sum of the forces in the y direction?
 
  • #5
3,812
92
That's where I am confused the net force of the x direction involves q but I have no idea how they want me to get that q. Also how would the trig ratio tan x/y help me in this case. They are asking for the field.

I guess you are given q (if you mean charge) in the question.
A diagram would help you to understand the situation:
(see attachment)
I rearranged the Fx in the second diagram.
So your next step is to find Fx, which is equal to the sum of the force exerted by the second sphere on the first sphere and the force due to electric field.
 

Attachments

  • tan.jpg
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  • #6
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Hmm so fy is mg, which means Fx is mgsin(theta) ?
 
  • #7
3,812
92
Hmm so fy is mg,
Correct! :smile:
which means Fx is mgsin(theta) ?
No, how did you get that? I already mentioned about Fx in my last post.
 
  • #8
327
1
Yea I think I figured it out!
Tension ends up canceling out
sum of the x direction ƩFx= Tsin(∅)=Eq

ƩFy=Tcos(∅)=mg

sin(∅)/cos∅=Eq/mg

Solving for E we get

E=mgtan∅/q

I hope that's right lol.
 
  • #9
3,812
92
Yea I think I figured it out!
Tension ends up canceling out
sum of the x direction ƩFx= Tsin(∅)=Eq

ƩFy=Tcos(∅)=mg

sin(∅)/cos∅=Eq/mg

Solving for E we get

E=mgtan∅/q

I hope that's right lol.

You are almost there but you missed one force which is the force exerted by the second sphere on the first sphere (which we are dealing with here). Calculate ƩFx again.
 
  • #10
327
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hmm honestly I cant think of it. They should be the same but different only difference is they are going in opposite directions. How would I include it?
 
  • #11
3,812
92
When there was no electric field, the only force on the left sphere in the x-direction was due to the the negatively charged right sphere. Direction of force being towards the right sphere because they attract each other. So Fx is equal to the force exerted by the right sphere on the left one. When the electric field is switched on, the positively charged left sphere experiences force in the direction of electric field. So, then Fx = FLR-qE, where FLR is the force on the left sphere by the right sphere.
 
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