E field magnitude

  • Thread starter sasuke07
  • Start date
  • #1
54
0

Homework Statement


What is the magnitude of the collective contribution to the E field at the origin from both the + and - charges located 5 cm above the horizontal axis?

Homework Equations


kQ/d^2



The Attempt at a Solution


Im just having a problem setting up this problem. Could someone help me on getting started
 

Attachments

  • charges.pdf
    8.2 KB · Views: 131

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
251
hi sasuke07! :smile:

(try using the X2 button just above the Reply box :wink:)

first, find the magnitude of each contribution,

from that, find the x and y components …

what do you get? :smile:
 
  • #3
54
0
I already figured out the magnitued of the the positive charge 5 cm above the axis and got 5.4X10^4. What do you mean by the x and y components. WOuldn't the negative charge have the same magnitude of the positive charge.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,836
251
WOuldn't the negative charge have the same magnitude of the positive charge.

yes :smile:
What do you mean by the x and y components.

draw an arrow at the origin showing the direction of the field from the positive charge

then find the x and y components of that arrow

(and finally you'll add them to the x and y components of the arrow from the negative charge :wink:)
 
  • #5
54
0
so to figure out the x component wouldn't it be Kq/d^2 where k=9X10^9, q is 30X10^9= charge and d would be .05m^2. Or would i have to use cosine and sine to figure out the x and y components and if so could you show me how to set it up.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,836
251
no, you must use cos and sin, of the angle the arrow makes
 
  • #7
54
0
so would it be Kq/r^2sintheta
and Kq/r^2Costheta. Where theta would be 90 degrees?
 
  • #8
54
0
or would theta be 45
 
  • #9
tiny-tim
Science Advisor
Homework Helper
25,836
251
θ is the angle of the line from the charge to the origin
 
  • #10
54
0
awesome so 45 degrees.
 
  • #11
54
0
So were the equations correct though?
 
  • #12
tiny-tim
Science Advisor
Homework Helper
25,836
251
looks ok :smile:

so what is the total field from both the positive and the negative charge?​
 
  • #13
54
0
i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions
 
  • #14
tiny-tim
Science Advisor
Homework Helper
25,836
251
i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions

your answer looks right to me :confused:

(are you sure it's 5 cm that they're asking about?)
 
  • #15
54
0
i think its 5cm because the length of the x and y components is 5cm.
 

Related Threads on E field magnitude

  • Last Post
2
Replies
28
Views
2K
Replies
2
Views
561
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
14
Views
4K
Replies
17
Views
188
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
9
Views
6K
Replies
4
Views
2K
  • Last Post
Replies
4
Views
776
Top