# Electric Work

1. Sep 27, 2008

### StephenDoty

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

#### Attached Files:

• ###### du.jpg
File size:
7.3 KB
Views:
43
2. Sep 27, 2008

### Staff: Mentor

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.

3. Sep 27, 2008

### StephenDoty

W=-delta*U

So what formula do I need to use?

4. Sep 27, 2008

### StephenDoty

the total potential energy=2kq^2/L

so would the work=-2kq^2/L?????

5. Sep 28, 2008

### Staff: Mentor

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.

6. Sep 29, 2009

### highcoughdrop

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?