Two point charges electric field

  • Thread starter kbyws37
  • Start date
  • #1
67
0
What are the magnitude and direction of the electric field midway between two point charges, −14.0 µC and +13.0 µC, which are 7.40 cm apart?

I used the F = ((k)(q1)(q2))/r^2

F= ((9x10^9)(-14)(13))/(7.40^2))

F= 2.99 x 10^10

however that answer is wrong. (The correct answer is 177000000 N/C).

Which equation am I suppose to use?
 

Answers and Replies

  • #2
hage567
Homework Helper
1,509
2
The equation you have there is for the force between two point charges, not the electric field due to a point charge. Take another look at your notes and see if you can figure out which equation it should be. You need to consider the force due to each charge at the centre, i.e. the electric field at the centre due to Q1 and the electric field at the centre due to Q2. It is a vector quantity. Make sure you are consistent with your units.
 
  • #3
67
0
The equation you have there is for the force between two point charges, not the electric field due to a point charge. Take another look at your notes and see if you can figure out which equation it should be. You need to consider the force due to each charge at the centre, i.e. the electric field at the centre due to Q1 and the force at the centre due to Q2. It is a vector quantity. Make sure you are consistent with your units.
So i would use the equation

F= ((k)(q))/r^2

so would i perform this equation twice? once for q1 and once for q2?

sorry, i am still not understanding
 
  • #4
hage567
Homework Helper
1,509
2
It isn't "F", it's "E".
Yes, that is the equation to use. You need to use it for both charges. The resultant electric field at the midpoint will be the vector sum of the fields from the two charges, E=E1+E2.
 
  • #5
67
0
It isn't "F", it's "E".
Yes, that is the equation to use. You need to use it for both charges. The resultant electric field at the midpoint will be the vector sum of the fields from the two charges, E=E1+E2.
I used the equation for both charges but still didn't get the right answer


E1 = ((9x10^9)(-14) / (7.4^2)) = - 2.3 x 10^9

E2 = ((9x10^9)(13) / (7.4^2)) = 2.14 x 10^9

E = (- 2.3 x 10^9) + (2.14 x 10^9) = 1.6

(correct answer should be 1.77 x 10^8)
 
  • #6
hage567
Homework Helper
1,509
2
You don't have the correct value for r. If the distance between the two charges is 7.4 cm, what is the distance from each charge to the midpoint of their separation?
Also, you are not being consistent with the magnitudes of the quantities. 14uC needs to be 14x10^-6 C, so your units work out. Do you see what I mean? Include your units in the calculation so you can make sure they work out properly. Check your conversion from cm to meters as well (you must do this).
 
  • #7
67
0
thank you! you explained it very well and i won't forget converting my numbers! :)
 
  • #8
hage567
Homework Helper
1,509
2
You should have all the information now to get the answer. Just put it all together in the right way!
 

Related Threads on Two point charges electric field

  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
6
Views
2K
Replies
3
Views
9K
  • Last Post
Replies
6
Views
8K
Replies
4
Views
1K
Replies
8
Views
58K
Replies
5
Views
3K
Replies
1
Views
1K
Replies
6
Views
7K
Replies
10
Views
4K
Top