Electric Field in a Square Problem

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field at point C in a square configuration due to charges A, B, and D. The user employs the formula E = kq/r² to determine the magnitude of the electric field, ultimately arriving at a final expression of 4√2 kq/a² for the electric field's magnitude. The user also discusses vector addition techniques, specifically the parallelogram rule, to combine electric field vectors. The direction of the electric field is established as West based on the vector diagrams drawn from the charges.

PREREQUISITES
  • Understanding of electric field concepts and vector addition
  • Familiarity with Coulomb's Law and the formula E = kq/r²
  • Knowledge of the Pythagorean Theorem and its application in physics
  • Basic skills in trigonometry, particularly with 45-degree angles
NEXT STEPS
  • Study vector addition techniques in physics, focusing on the parallelogram rule
  • Learn about the superposition principle in electric fields
  • Explore the implications of charge configurations on electric field direction and magnitude
  • Investigate the effects of different charge distributions on electric field calculations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in mastering electric field calculations in square charge configurations.

Kirasagi
Messages
9
Reaction score
0

Homework Statement



I think it would be better if I put the picture.

nTzgs.png


Additional questions:
Find the direction of the electric field.
Find the magnitude of electric field due at C due to charges A, B, and D.

Oq2YY.png

Homework Equations



Pythagorean Theorem.
E = kq/r^2

The Attempt at a Solution



For the first problem, I drew vector diagrams based on the charges at the center of the square. I ended up having 4 arrows:
O -> A (electron attraction)
O -> D (electron attraction)
O -> A (proton repulsion of charge C)
O -> D (proton repulsion of charge D)
So basically I had 2 visual arrows of pointing NW and SW from the origin. This makes the direction of the electric field to be West. Thats basically the part I only got.
For the magnitude of the electric field I used E = kq/r^2 . I know r = [√(2)/2]a. Now I got lost on what to really do (I tried a bunch of stuff but it didn't really make sense to me). The final answer for the first question is 4√2 kq/a^2
 
Physics news on Phys.org
Do you know how to sum vectors? eg Work out the x and y component of each vector. Sum all the x components and sum all the y components etc

To further simplify the equations it might help to remember that:

This is a square so 45 degree angles are involved.
cos(45)=sin(45)=0.5*SQRT(2)
Pythagorous involves squares and square roots

sorry if you know all that.
 
Thanks for the reply!

All along I was doing my math wrong on adding the electric field vectors.

As r = (sqrt2/2)a

e = kq/(r^2)
e = kq/(a^2/2)
e = 2kq/(a^2) for each vector on one side.

Since there's 2 vectors on each side then it equals to 4kq/(a^2)

Combining the vectors using parallelogram rule I get 4sqrt2 kq/(a^2)

For the third question (Find the magnitude of electric field due at C due to charges A, B, and D):
Do I just solve this problem similar to the 1st question? (draw vectors based on the center O). I drew the vectors due to C and here's what I came up with:

C -> A
C -> D
B -> down

Not sure if I'm starting this correctly. Do I just add the electric field vectors again based on the superposition rule? If so how would I do the vector pointing north from B (not sure what I would put for as r).
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 32 ·
2
Replies
32
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
11
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K