Magnitude of Electric field using superposition principle

Click For Summary

Homework Help Overview

The problem involves calculating the magnitude of the electric field at a point P due to two charges, +q and -q, positioned in the x-y plane. The original poster attempts to apply the superposition principle and relevant equations, but encounters difficulties in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the breakdown of the electric field components and question the distances used in the calculations. There is an exploration of the correct hypotenuse calculation for the distance from the charges to point P.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing feedback on the setup and calculations. Some guidance has been offered regarding the distance calculation and the consideration of electric field contributions from both charges. Multiple interpretations of the problem setup are being explored.

Contextual Notes

There are indications of potential misunderstandings regarding the geometry of the problem and the application of trigonometric functions in the calculations. The original poster's use of specific values and assumptions is being scrutinized.

skibum143
Messages
112
Reaction score
0

Homework Statement



Two charges, +q and -q, are located in the x-y plane at points (0,+d/2) and (0,-d/2), respectively. Calculate the magnitude of the electric field at point P with the superposition principle.
Data: q=37.0 nanoCoulombs,
d=4.60mm and P at
x=92.0mm.

Homework Equations


E = kQ/r^2
a^2 + b^2 = c^2
sin = o/h


The Attempt at a Solution


I tried to solve this like a normal E field question.
For the +q charge, I got that the E field would be:
( k * -37E-6 (sin2) / .092^2) + ( k * 37E-6 (cos2)
and For the -q charge, I got that the E field would be:
( ( k * -37E-6 (sin2) / .092^2) - ( k * 37E-6 (cos2)
This came out to 2740 N/C (positive b/c it asks for magnitude) but that was wrong. Can someone tell me where I'm going wrong? I attached the diagram.
 

Attachments

  • prob11a_dipole1.gif
    prob11a_dipole1.gif
    2.5 KB · Views: 824
Physics news on Phys.org
I understood what you first did,

you first said that you will break up E= Esintheta + Ecostheta, one thing I noticed that in the equation E=kq/r^2 , you used for r = 0.092 you used the x distance for the point p, I suggest you have another look at your figure :

http://img231.imageshack.us/img231/5466/59813260.gif

compare the blue line with the red one , do you still think that the distance between q and p is x (0.092) ?
 
Last edited by a moderator:
When I did (.092^2) + (.0023^2), I get that the hypotenuse is .0092...
 
Ah, sorry my mistake, I now get .00847. I will try that. Was the rest of the setup ok? Thanks for your help!
 
Hmmm... that was still wrong. Am I doing something else incorrectly?
 
i think the required distance between q and p should be sqrt((0.092^2) + (0.0023^2)) i did the calculation and i got 0.092028 (when you said it is 0.00847 you didnt take the square root) ..

another thing, I think you probably haven't considered the elecrtic field produced from positive charge and negative charge, have a look at this figure:

http://img402.imageshack.us/img402/7681/95152182.jpg

so your problem should look something like the following:

http://img4.imageshack.us/img4/4996/34194234.gif

notice where r the electric field vectors , find the components of each vectors ,then see whether you will find any compenents that will cancel with each other and pay attention to the directions ..

by the way , i have one question for you , from where you get sin2 and cos2?
 
Last edited by a moderator:

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
4K