Determine the point at which the electric field is zero

Click For Summary
SUMMARY

The discussion centers on determining the point at which the electric field is zero between two charges: -2.50 C and 6.00 C, positioned 1.00 m apart. Participants utilized the formula E = kq/r² to calculate the electric fields (E1 and E2) and set E1 + E2 = 0 to find the equilibrium point. The correct solution is 1.82 m to the left of the negative charge, as confirmed by multiple users after correcting calculation errors. The book referenced is "Physics for Scientists and Engineers with Modern Physics, Volume 2" by Serway and Beichner.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with quadratic equations and their solutions
  • Knowledge of the concept of superposition in electric fields
  • Basic proficiency in algebraic manipulation and solving equations
NEXT STEPS
  • Study the principles of electric field superposition in electrostatics
  • Learn how to apply the quadratic formula effectively in physics problems
  • Explore the concepts of electric potential and its relationship to electric fields
  • Review problem-solving techniques for electrostatic scenarios in "Physics for Scientists and Engineers" by Serway and Beichner
USEFUL FOR

Students and educators in physics, particularly those focusing on electrostatics, as well as anyone seeking to improve their problem-solving skills in electric field calculations.

oooride
Messages
34
Reaction score
0
Two charges -2.50 C and 6.00 C are 1.00 m apart, determine the point at which the electric field is zero.

I thought this problem would be easy but I must be doing something ridiculously wrong because I've been stuck on this problem for a about an hour...

I broke up the problem into E1 & E2 using E = kq/r^2, and replacing r^2 in E1 with x^2 and r^2 in E2 with (x+1.00m)^2 and tried to solve using the quadratic formula without success..

Any hints?

thanks in advance :)
 
Physics news on Phys.org
Everything you did is correct.
You use E1+E2=0, use your xlike you said and
solve. Try it again.
(I get about 1.91 m)
 
hmmmm..

It's been a long day so maybe I'm making some stupid mistake, but I'm still unable to come up with anything near the answer.. The book shows (1.82m to the left of the negative charge) as the answer, and you're at ~1.91m, very much closer than my answer.


Here's what I did:

E1=KQ/x^2
E2=KQ/(x+1)^2

E_net = E1 + E2

E_net = 6KQ/x^2 - 2.50KQ/(x+1.00)^2

2.50/(x+1.00m)^2 = 6/x^2

2.50x^2 = 6(x+1.00m)^2

2.50x^2 = 6x^2 + 12.00x + 6.00

6x^2 - 2.50x^2 + 12.00x + 6.00 = 0

3.50x^2 + 12.00x + 6.00 = 0

-12.00 +- sqrt (12.00)^2 - 4(3.50)(6.00) / 2(3.50)

-12.00 +- sqrt (144)-(84) / 7.00

-12.00 +- 7.75 / 7.00

= -0.607 and -2.82 which I believe is completely wrong...


Am I making a stupid calc error? Am I not finishing the problem? or did I just completely approach it wrong?

Thanks for the hints in advance.. :)
 
Oops... your book is correct I made a calc mistake
by getting delta = sqrt(100+180) and it's supposed
to be sqrt(100+140). Which gives you 1.82.

Anyway, the electric field will be 0 farther (1+x) from the
6 C than from the -2.5 C (x) so you've simply put
the distances in reverse in your field strenght equations.

Live long and prosper.
 
Ahhhhhh... I came out with 1.82m now. Thanks!
 
oooride,
I had that exact problem to work out too. We must be using the same book, "Physics for Scientist and Engineers", 6th edition by Serway and Jewett?

I got stuck on this one until the prof went over it in class.
After you posted it here, I went through to see if I can solve it on my own, and I GOT STUCK AGAIN!

Thanks for the review reminder.
 
Originally posted by paul11273
oooride,
I had that exact problem to work out too. We must be using the same book, "Physics for Scientist and Engineers", 6th edition by Serway and Jewett?

I got stuck on this one until the prof went over it in class.
After you posted it here, I went through to see if I can solve it on my own, and I GOT STUCK AGAIN!

Thanks for the review reminder.


Yup that's the book, except I'm using the fifth edition instead of the sixth.

The book's title is "Physics for Scientists and Engineers with Modern Physics, Volume 2, by Serway and Beichner".
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K