Calculating Force Between Two Conducting Spheres

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Homework Help Overview

The problem involves calculating the electrostatic force between two conducting spheres with given charges and separation distance. It also explores the scenario where the spheres are touched together and then separated again.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Coulomb's law to find the force between the spheres and question the calculation of the force, particularly the sign and magnitude. There are attempts to understand the effect of touching the spheres on their charges and the resulting force.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and reasoning. Some have provided insights into the charge redistribution when the spheres touch, while others are questioning the initial calculations and the implications of negative force values.

Contextual Notes

Participants note the importance of understanding the direction of forces due to the nature of the charges involved. There is mention of homework constraints, such as the requirement to explore the effects of touching the spheres on their charges.

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Homework Statement



Two conducting spheres of the same size have charges q1=-4x10^-6
and q2 = 8x10^-6C,and their centers are separated by 3.0cm

a) what is the force between them?
b) what would be the force if the spheres were touched together and again separated by the same distance

Homework Equations



well we could use F=kq1q2/r^2


The Attempt at a Solution




for a) I plugged in 9x10^9 x -4x10^-6 x 8x10^-6 all divided by .09 which gives -3.2 but the answer is -320N.can someone explain my why?

for b) i don't know how it would start.
i tried doing several things. use the -320 as the separated force and
get the force of the one together and add them together.

something like that but i don't know how it gives 40N.
 
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any help.
i still can';t figure this basic problem out
 
Well... for a
F = kq1q2/r^2.. like you said. I plugged it in my calculator, and did
(9x10^9)(4x10^-6)(8x10^-6)/(.03)^2 and got 320 N... perhaps you did a calculation error, but that is how you set it up. I don't remember in class directly putting the negative into the equation, but perhaps the force is negative because you have opposite charges? (I'm not sure on direction)

As for b, if both spheres touch each other, afterwards they must have the same charge. If the spheres touch, then they will receive the same charge (2 x 10^-6 C). Therefore, just apply the new values of q to the equation.. and you should get something like
F = kq1q2/r^2 = (9x10^9)(2x10^-6)(2x10^-6)/(.03)^2 = 40 N

so a is a) 320 N (not sure about direction) and b is 40 N (also not sure of direction) ... do you see why?
 
Oh.. and the reason the spheres each receive 2 x 10^-6 charge is because that is the amount of charge that is in the middle of both charges... notice how the first has -4 x 10^-6 and the second has 8 x 10^-6.. well the middle of -4 and 8 is 2, so that's how much each charge has! Just remember that when two spheres touch each other (of opposite charge), they must end up with the same charge.. so just take the middle of it!
 
thanks for the great explanation.

the directions are just attractice and repulsive respectively.
 
Oh okay ^^. You're welcome.. I hope it helped!
 

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