Electric potential due to collection of charged particles

Click For Summary

Homework Help Overview

The problem involves calculating the electric potential at the origin due to two charged particles located at specified coordinates. The subject area pertains to electrostatics and electric potential calculations.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for electric potential and attempt to substitute values for the charges and distances. There are questions regarding the accuracy of the calculations and the values used in the computations.

Discussion Status

Some participants have expressed confusion over the calculations, while others have pointed out potential errors in the distance measurements used. There is a recognition of the challenges in performing the calculations accurately.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can share or the methods they can use. There is an indication of differing interpretations of the distances involved in the problem.

Rijad Hadzic
Messages
321
Reaction score
20

Homework Statement


A particle with charge q1 = -6.75x10^-6 is located at (0,3.25x10^-2 m ) and a second particle with charge 3.2x10^-6 is located at (0,-2.75x10^-2 m) what is the electric potential due to the two charges at the origin?

Homework Equations


V = kq/r

The Attempt at a Solution



So at the origin the potential should be kq1/r1 + kq2/r2 = k(q1/r1 + q2/r2)

plugging in, (8.99x10^9)(-1733785 + 1046109) = -6.87x10^5

but my book is telling me the answer is -8.21x10^5

what am I not taking into account here?
 
Physics news on Phys.org
Rijad Hadzic said:
So at the origin the potential should be kq1/r1 + kq2/r2 = k(q1/r1 + q2/r2)
Good.

plugging in, (8.99x10^9)(-1733785 + 1046109)
I don't get the values shown in red.
 
TSny said:
Good.

I don't get the values shown in red.

Strange. I did the calculation a million times exactly like I just did it but I got the answer now... the mechanical task of calculating stuff makes me want to pull out my hair. I hope I don't do this on a test lol.

Anyways thank you..
 
OK, good. For the first charge, it looks like you used a distance of .035 m rather than .0325 m.
 

Similar threads

Replies
5
Views
1K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
6
Views
2K
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K