Electrostatics: Two charged balls with attraction

In summary, the first problem involves two fixed conducting balls with known charges and a distance between them. After connecting them with a wire and then removing it, the balls repel each other with a known force, allowing us to solve for the charge of the second ball and the separation between the balls. The second problem involves finding the position between two particles where a third particle experiences no net force, using the formula F=k(q1+q2)^2/(4*r^2) and solving a quadratic equation.
  • #1
thebe39
5
0
Hi, I have an exam coming up soon. But I thought I would clear my doubts ahead of time.

1. A fixed conducting ball has charge q1= 3X10^-6C. An identical ball with charge q2 is held at a distance x away from q1. The two balls attract each other with a force of 13.5N. The balls are then connected by a conducting wire. After the wire is removed, the balls repel each other with a force of 0.9N.

a) what was the charge of q2 of the second ball?
b) What is the separation x between the balls?


2. At what position between particles 1 and 2 will particle 3 experience no net force?
q1= 2x10-9 q2=3x10-9 q3= -2x10-9. The distance between particle 1 and 2 is 0.04.


Thanks
 
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  • #2


Welcome to PF.

Where did you get stuck on the first problem?
 
  • #3


well I know that in the beginning the q2 should have a - charge since it is attracted to the q1 and then after the wire it should become -. I'm not really sure how to isolate for
q2 from the equation.
 
  • #4


thebe39 said:
well I know that in the beginning the q2 should have a - charge since it is attracted to the q1 and then after the wire it should become -. I'm not really sure how to isolate for
q2 from the equation.

If the charge from both is the same it will be repulsive.

That means in the first you a force made up of the given charge times the unknown. The second force is derived from the average of the 2 charges squared.

F = kq1*q2/r2

After

F = k(q1 + q2)2/(4*r2)

Where (q1 + q2) / 2 is the average that gets squared.
 
  • #5


how did you get F=Kq(q1+q2)/(4*r^2) ? where does the extra q come from?
 
  • #6


btw the answer to this problem is apparently
a) -5x10-6 C
b) 0.10m

But I really just want to understand what I'm doing with the steps.
 
  • #7


thebe39 said:
how did you get F=Kq(q1+q2)/(4*r^2) ? where does the extra q come from?

Sorry, there is no extra q. I noticed the typo and edited it apparently after you picked up the post.
Note that the correct formula is

F = k(q1 + q2)2/(4*r2)
 
  • #8


So I'm going to let r^2 equal each other

kq1q2/F = k(q1 +q2)^2/F(4)

So I isolate for q2 right?

or do I say that kq1q2/F = k(q1 +q2 divide by 2)^2/F(4)
 
  • #9


thebe39 said:
So I'm going to let r^2 equal each other

kq1q2/F = k(q1 +q2)^2/F(4)

So I isolate for q2 right?

or do I say that kq1q2/F = k(q1 +q2 divide by 2)^2/F(4)

The factor of 4 came from q=q1+q2 divide 2, so your first equation is correct, but not your second. Note that the two "F"'s are different: in kq1q2/F, F=-13.5 N, but in k(q1+q2)^2/4F, F=0.9 N.

Otherwise, you're on the right track. Cancel out the k, expand the right side, and you'll get a quadratic equation. It should be easy from there. Remember that q1, F, and the other F are all given.
 

Related to Electrostatics: Two charged balls with attraction

1. What is electrostatics and how does it relate to two charged balls with attraction?

Electrostatics is the study of electric charges at rest. In this case, two charged balls with attraction refers to two objects with opposite charges that are attracted to each other due to the presence of these charges.

2. What factors affect the strength of the attraction between two charged balls?

The strength of attraction between two charged balls depends on the magnitude of the charges on each ball and the distance between them. The greater the magnitude of the charges and the closer the distance between the balls, the stronger the attraction.

3. How is the attraction between two charged balls calculated?

The attraction between two charged balls can be calculated using Coulomb's Law, which states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

4. Can two negatively charged balls attract each other?

Yes, two negatively charged balls can attract each other. This is because the charges on the balls are opposite in sign, and according to Coulomb's Law, opposite charges attract each other.

5. How does the presence of other objects or charges affect the attraction between two charged balls?

The presence of other objects or charges can affect the attraction between two charged balls. For example, if there are other positively charged objects nearby, they may interfere with the attraction between the two charged balls by creating an electric field that opposes the attraction. Additionally, if there are other negatively charged objects nearby, they may attract one of the charged balls, reducing the overall attraction between the two balls.

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