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Plastic Balls in a Test Tube with Mass

  • #1
191
0

Homework Statement



One charged plastic ball is vertically above another in a test tube as shown. The balls are in equilibrium a distance "d" apart. If the mass of the upper ball is reduced by a factor of four, the distance between the balls in the test tube would become :

1) 4d
2) 8d
3) 2d
4) square root of 2d

Homework Equations



?

The Attempt at a Solution



How are charge and mass related? That's what I can't understand.
 

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Answers and Replies

  • #2
rl.bhat
Homework Helper
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Charge decides the repulsive force and mass decides the gravitational attractive force. In equilibrium position they are equal amid opposite.
 
  • #3
191
0
Ok. I'm thinking along the lines of

F initial = G m1m2 / r^2

F final = G 1/4m1m2/ r^2

But I need to put those in terms of "r" and compare, right?
 
  • #4
rl.bhat
Homework Helper
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By changing mass of tthe plastic ball, the weight mg changes. Now compare the two weights with the force of repulsion.
 
Last edited:
  • #5
191
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Ok. What I just did was

Gm1m2
--------
1/4 Gm1m2

to get a 4:1 ratio. Yet I didn't use anything with "R."
 
  • #6
rl.bhat
Homework Helper
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When mass decreases, the weight decreases. To balance this the charges must move farther. What is the formula for the force of repulsion?
 
  • #7
191
0
F = kq1q2/r^2 ?
 
  • #8
rl.bhat
Homework Helper
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Curret. Now if the initial distance for equilibrium is d for mg , what is new distance to balance mg/4?
 
  • #9
191
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I'm thinking:

d initial = mg
--------------
d final 1/4 mg

solving for d initial I get:

d initial = 4d final

But there's a square involved so shouldn't the answer be "2d?"
 
  • #10
rl.bhat
Homework Helper
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5
Yes.
 

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