Electric force between two spheres?

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Homework Help Overview

The discussion revolves around the electric force between two charged spheres and the effects of changing charge and distance on that force. Additionally, there is a parallel discussion on the resistance of wires and the behavior of current in parallel resistor circuits.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between charge, distance, and electric force, with some asserting that doubling one charge would increase the force. Others analyze the impact of changing resistance in parallel circuits and question the effects of increasing resistance on current flow.

Discussion Status

Some participants provide insights into the relationships governing electric force and resistance, while others express confusion about the implications of increasing resistance in parallel circuits. There is a mix of agreement on certain points, but also ongoing questions and clarifications being sought.

Contextual Notes

Participants are navigating concepts of electric force and resistance, with some assumptions about the nature of the problems being discussed. The original poster's question about electric force is distinct from the subsequent discussion on resistance, indicating a shift in focus within the thread.

UrbanXrisis
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Two identical small spheres possessing charges q1 and q2 are separated by distance r. Which charge would produce the greatest inecreas in the electric force between the tow spheres?

1. double charge q1
2. double r
3. double r and q1
4. double r, q1, and q2

I'm pretty sure it's #1
 
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Since force goes as product of charges and inverse square of distance,

#1 doubles force
#2 cuts force to 1/4 of what it was
#3 cuts force to 1/2 of what it was
#4 leaves force the same

So you are correct in answering #1.
 
The resistance of a wire at constant temperature depends on the wire's...length and cross sectional area. Does the type of meteal come into play?
 
Of course it does.

Look up resistivity.
 
ohhh, right R=resistivity*L/A

In a parallel connection of two resistors, if the resistance of the resistors were increased, the current would decrease right? According to R=V/I
 
You get a *qualified* yes there.

Yes, if you increase the resistance the current decreases. But remember that if the resistors are in parallel you can't simply add them together, right?
 
In a parallel, if the total resistance decreases when the resistors increase in resistance dure to 1/Rt=1/R1+1/R2...

So, if resistance in the resistors were to increase in a parallel, then really, the total resistance would decrease making the current increase??

I'm confused now
 
2 resistors in parallel: r1 = 4; r2 = 5
\frac{1}{4} + \frac{1}{5} = \frac{9}{20} \textrm{giving} R_T = \frac{20}{9}

increase: now r2=5
\frac{1}{5} + \frac{1}{5} = \frac{2}{5} \textrm{giving} R_T = \frac{5}{2}

5/2 is greater than 20/9

OK?

Effective total resistance has increased, so the current will decrease.
 
Increasing R1 would decrease 1/R1. Increasing R2 would decrease 1/R2. Doing either one of those will decrease the right side, and therefore the left side must also decrease. But that means the reciprocal of the left side would increase, and the reciprocal is in fact the total resistance, so total resistance goes up.
 
  • #10
Gnome beat me to the answer.
 
  • #11
Okay, but the more resistors you add, the less the total resistance right?
 
  • #12
That is true, as long as you add them in parallel with the other ones.
 
  • #13
thanks guys, that cleared up a lot of things
 
  • #14
Urban, do you realize the difference between increasing the resistance of an existing resistor vs. adding an additional resistor in parallel with the others?

In the first instance, you are further restricting the flow of current; the combined effective resistance is increased.

In the second, you are actually adding an additional pathway for the current to flow through, so the combined effective resistance is decreased.
 

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