Electric force between two spheres?

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SUMMARY

The discussion centers on the electric force between two charged spheres and the effects of varying charge and distance on that force. Doubling the charge of one sphere (option #1) results in the greatest increase in electric force, while doubling the distance (option #2) reduces the force to one-fourth. The conversation also touches on the principles of resistance in electrical circuits, specifically how increasing resistance in parallel circuits affects total resistance and current flow. Key formulas discussed include the electric force equation and the resistance formula R = V/I.

PREREQUISITES
  • Understanding of Coulomb's Law for electric force calculations
  • Knowledge of Ohm's Law (R = V/I) for analyzing electrical circuits
  • Familiarity with the concept of resistivity in materials
  • Basic principles of series and parallel resistor configurations
NEXT STEPS
  • Study Coulomb's Law and its applications in electrostatics
  • Learn about the concept of resistivity and its impact on electrical resistance
  • Explore the differences between series and parallel circuits in detail
  • Investigate the implications of varying resistance on current flow in electrical systems
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone interested in understanding the principles of electric forces and circuit behavior.

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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