Work needed to move opposite charges farther apart

  • Thread starter mikil100
  • Start date
  • #1
12
0

Homework Statement


Suppose you want to move two charges apart, both are equal charges of opposite sign at 1.4x10^9C, and their starting position is 6.7m apart, how much work would it take to further separate the charges to a final distance of 7.9m, that is, move them 1.2m

Homework Equations


W=Kq1q2/r[/B]


The Attempt at a Solution


[/B]
I have gotten an answer at 1.5x20^28 J, this was by taking the distance 1.2m and using it as the distance traveled-plugging it into the equation-- I also made an assumption that both charges, while opposite sign are treated with absolute value, otherwise I get negative work which doesn't seem to make sense as it is working to separate the charges from on another, therefore increasing potential energy.
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
Hi mikil100, Welcome to Physics Forums.

Your Relevant equation is not quite right. As it stands it doesn't express work, but potential energy in the configuration of two charges q1 and q2 that are a distance r apart. So your answer that came from considering r to be the change in separation is not right.

So a valid approach for you would be to consider the potential energy at the initial distance and the potential energy at the final distance. The difference between them would be the work done to move from one position to the other.

Alternatively you could consider the electrostatic force between the charges and do the Force x distance integration over the path from the initial separation to the final separation.

Your observation on dealing with the signs of the charges is a good one. When you're contending with charge signs affecting the force directions and choices of coordinate system axis directions, it can get tricky to juggle all the parameters of the problem. It's always a good idea to draw a sketch and determine the directions of the forces and their relationships to any movement made. Then go ahead and use absolute values for the charges and use signs that you determined from your sketch on the terms to specify the directions.
 
  • #3
12
0
So I could simply calculate the distance traveled, being 1.2m by multiplying it by the force interacting between the two charges ( using the formula F=k(q1q2)/r^2?

Using W=F*d I get 4.68x10^26

If I use the other method, being position initial-position final (which would have lower PE) , it would be (k(1.4x10^9)^2/6.7) - (k(1.4x10^9)^2/7.9) equalling 4.3x10^26

Does this seem equatable? Thank you for your help!
 
  • #4
gneill
Mentor
20,925
2,867
So I could simply calculate the distance traveled, being 1.2m by multiplying it by the force interacting between the two charges ( using the formula F=k(q1q2)/r^2?

Using W=F*d I get 4.68x10^26
Nope. Because the force changes with distance. That's why I mentioned doing the integration over the path.
If I use the other method, being position initial-position final (which would have lower PE) , it would be (k(1.4x10^9)^2/6.7) - (k(1.4x10^9)^2/7.9) equalling 4.3x10^26
You may want to go back to your calculator there. I'm not seeing that value as a result with those figures punched in. What value are you using for k?
 
  • #5
12
0
Nope. Because the force changes with distance. That's why I mentioned doing the integration over the path.

You may want to go back to your calculator there. I'm not seeing that value as a result with those figures punched in. What value are you using for k?
Ah, okay... that makes sense, unfortunately this is an algebra based class I am taking and my own personal ability at integrating is limited.

I am using Coulombs constant for k, at 9x10^9.
I rechecked my math and I have rounding error, carrying the equation through with my calculator I get a value of 3.99x10^26. Does this seem more accurate?
 
  • #6
gneill
Mentor
20,925
2,867
I rechecked my math and I have rounding error, carrying the equation through with my calculator I get a value of 3.99x10^26. Does this seem more accurate?
Yup! Much better. Be sure to specify the units when you submit your answer!
 

Related Threads on Work needed to move opposite charges farther apart

  • Last Post
Replies
2
Views
20K
  • Last Post
Replies
1
Views
2K
Replies
14
Views
977
  • Last Post
Replies
5
Views
12K
Replies
4
Views
4K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
15K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
3K
Top