# Homework Help: Electric Field Problem-Balancing Forces

1. Feb 2, 2013

### tyreal

1. The problem statement, all variables and given/known data

A 0.020g plastic bead hangs from a lightweight thread. Another bead is fixed in position beneath the point where the thread is tied. If both beads have charge q, the moveable bead swings out to the position shown (attachment). What is q?

2. Relevant equations

F=[K(q1)(q2)]/r^2
w=mg

3. The attempt at a solution

I am new to physics forums so please excuse my equations formatting! I know this is a force balancing problem. I think that tension force of string equals weight force of bead. There is a repulsive electrical force between the two beads, which I think equals the tension/gravitational force that is holding the bead in place.

T=mg=0.000196N

My question is, should I set this equal to the expression of [K(q1)(q2)]/r^2 and solve for q? I know that is the last step since I am trying to find q. BUT, my professor said we had to use sin45 or cos45 in some way so as to separate the x and y components of the force. I don't see where to incorporate that. I am missing some sort of middle step here, any input would be appreciated!!

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2. Feb 2, 2013

### G01

Hi tyreal! Welcome to PF.

Remember when we write force balancing equations we want to balance the forces in both x and y directions separately.

Hint: The force of gravity only has a y component. Thus, the force of gravity must be balanced out by the y component of the tension force. Similarly, in this problem, the electric force has only an x-component. Thus, the electric force is balanced by the x component of the tension.

Can you write two force balancing equations expressing the above?

3. Feb 2, 2013

### tyreal

Okay, that helped a lot. Here is what I got when I separated the two:

Fx(electric)=Tsin45=[K(q1)(q2)]/r^2
Fy(gravity)=Tcos45=mg

Can I assume T=mg for both the x and y cases? Then I can ignore the Y component since the question is only asking about forces that are on the X axis.

Solving mg*sin45=[K(q1)(q2)]/r^2, using 9*10^9 for K, 0.05m for r, 0.00002kg for m and 9.8 for g, I get q=6.20*10^-9 C, which seems appropriate.

4. Feb 2, 2013

### haruspex

But you have just shown that Tcos45=mg.

5. Feb 2, 2013

### G01

haruspex is correct. If the rope was hanging straight down, T=mg, but that is not the case in this problem because of the electric repulsion.

Use the two equations you have to solve for q.

6. Feb 3, 2013

### tyreal

Very good input here! Thank you for pointing that out haruspex. If I cannot assume T=mg for the horizontal forces (supported by commonsense reasoning!!), what would I use for T? And what would I be left with to set equal to [K(q1)(q2)]/r^2? G01, I am not sure which two equations you are referring to.

7. Feb 3, 2013

### G01

These two:
You have two equations and two unknowns, T and q. You should be solve the equations simultaneously and solve for both T and q.