# Plastic Balls in a Test Tube with Mass

## 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

?

## The Attempt at a Solution

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

#### Attachments

• 6.8 KB Views: 353

Related Introductory Physics Homework Help News on Phys.org
rl.bhat
Homework Helper
Charge decides the repulsive force and mass decides the gravitational attractive force. In equilibrium position they are equal amid opposite.

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

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

F = kq1q2/r^2 ?

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

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
Homework Helper
Yes.