Electric Field Problem-Balancing Forces

In summary: Then you can use one of the equations to solve for q.In summary, for a 0.020g plastic bead hanging from a lightweight thread, with another fixed bead below it, both with charge q, the moveable bead swings out to a certain position. By balancing the forces in the x and y directions using the equations Fx(electric)=Tsin45=[K(q1)(q2)]/r^2 and Fy(gravity)=Tcos45=mg, we can solve for both T and q. Using the values given, we can determine that q=6.20*10^-9 C.
  • #1
tyreal
3
0

Homework Statement



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?

Homework Equations



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

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
tyreal said:

Homework Statement



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?

Homework Equations



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

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!

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
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
tyreal said:
Fx(electric)=Tsin45=[K(q1)(q2)]/r^2
Fy(gravity)=Tcos45=mg
Can I assume T=mg for both the x and y cases?
But you have just shown that Tcos45=mg.
 
  • #5
tyreal said:
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.

haruspex said:
But you have just shown that Tcos45=mg.

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
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
tyreal said:
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.
These two:
tyreal said:
Tsin45=[K(q1)(q2)]/r^2
Tcos45=mg

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

FAQ: Electric Field Problem-Balancing Forces

1. What is an electric field?

An electric field is a force field that surrounds an electrically charged particle. It is a vector quantity, meaning it has both magnitude and direction.

2. How do you calculate the electric field?

The electric field is calculated by dividing the force exerted on a test charge by the magnitude of the test charge. It is also influenced by the distance between the charged particles and the type of medium in which they are located.

3. What is the principle of balancing forces in an electric field?

The principle of balancing forces in an electric field states that when two charged particles are placed in an electric field, the forces acting on them will be equal and opposite in direction, resulting in a state of equilibrium.

4. How do you determine the direction of the electric field?

The direction of the electric field is determined by the direction of the force exerted on a positive test charge. If the force is in the same direction as the electric field, the charge will experience a repulsive force. If the force is in the opposite direction, the charge will experience an attractive force.

5. What are some real-world applications of balancing forces in an electric field?

Balancing forces in an electric field is essential in many modern technologies, such as capacitors, electric motors, and generators. It is also crucial in understanding the behavior of charged particles in the Earth's atmosphere, which is essential in predicting and preventing lightning strikes.

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