Plastic Balls in a Test Tube with Mass

[PeachBanana]

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.

 

[rl.bhat]

Charge decides the repulsive force and mass decides the gravitational attractive force. In equilibrium position they are equal amid opposite.

 

[PeachBanana]

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?

 

[rl.bhat]

By changing mass of tthe plastic ball, the weight mg changes. Now compare the two weights with the force of repulsion.

 

[PeachBanana]

Ok. What I just did was

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

to get a 4:1 ratio. Yet I didn't use anything with "R."

 

[rl.bhat]

When mass decreases, the weight decreases. To balance this the charges must move farther. What is the formula for the force of repulsion?

 

[PeachBanana]

F = kq1q2/r^2 ?

 

[rl.bhat]

Curret. Now if the initial distance for equilibrium is d for mg , what is new distance to balance mg/4?

 

[PeachBanana]

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?"

 

[rl.bhat]

Yes.