Calculate W for Point Charge Configuration

AI Thread Summary
To calculate the work W for a point charge configuration, the contribution from each charge is proportional to kq^2/L. The discussion highlights that the work done must account for all charge pairs, not just individual contributions, leading to the formula W = -2kq^2/L for total work. The potential energy between two charges is given by kq_1q_2/r, and the total potential energy for multiple charges is the sum of each pair's energy. It is emphasized that the work is negative due to the convention of potential energy being zero at infinity. Understanding the contributions from all charge pairs is crucial for accurate calculations.
StephenDoty
Messages
261
Reaction score
0
If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinitely far apart, you will find that the contribution for each charge is proportional to {kq^2}/{L}. In the space provided, enter the numeric value that multiplies the above factor, in W. (See Picture Below)

delta U= -W
delta U= q*delta V

I tried using this idea but my answers are wrong. Like for charge A I got delta V= kq/sqrt(2)*L
or delta U= kq^2/sqrt(2)*L thus W = -kq^2/sqrt(2)*L

But the work for charge A equals 0. So what am I doing wrong?

Thanks
Stephen
 

Attachments

  • du.jpg
    du.jpg
    4.5 KB · Views: 380
Physics news on Phys.org
StephenDoty said:
I tried using this idea but my answers are wrong. Like for charge A I got delta V= kq/sqrt(2)*L
or delta U= kq^2/sqrt(2)*L thus W = -kq^2/sqrt(2)*L
This only considers the charge pairs that include A. (Why the negative sign?)

To find the total work done you must consider every pair of charges: A-B, A-C, A-D, B-C, and so on.
 
W=-delta*U

So what formula do I need to use?
 
the total potential energy=2kq^2/L

so would the work=-2kq^2/L?
 
StephenDoty said:
the total potential energy=2kq^2/L

so would the work=-2kq^2/L?
How did you determine this value for PE?

As I'm sure you know, the potential energy between two charges separated by a distance r = kq_1q_2/r. Not that the PE is zero at infinity, thus the work done to move these two particles from infinity to a distance r is just kq_1q_2/r (no need for a negative sign).

To find the total potential energy for all four particles, add up the potential energy contribution of each pair of charges. List each distinct pair (there are six) and its potential energy.
 
http://www.physics.umd.edu/courses/Phys260/agashe/S09/solutions/HW11.pdf
 
Last edited by a moderator:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top