# Homework Help: Four Equidistant Particles in electric fields

1. Feb 4, 2012

### C6ZR1

1. The problem statement, all variables and given/known data
In Fig.22-30, the four particles form a square of edge length a = 6.50 cm and have charges q1 = 7.59 nC, q2 = -10.9 nC, q3 = 11.5 nC, and q4 = -6.06 nC. What is the magnitude of the net electric field produced by the particles at the square's center?

http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c22/fig22_32.gif

2. Relevant equations

E=kq/r^2

3. The attempt at a solution

First thing, I converted charges from nC to C as well as distances from cm into m

q1= 7.59 E-9 C
q2= -1.09 E-8 C
q3= 1.15 E-8 C
q4= -6.06 E-9 C

a=0.065m

after that, I formed triangles and used Pythagorean theorem to get distance for each particle to the center.

Center distance (hypotenuse) = 0.045962

So what I thought was find the Electric field produced by each particle with r= the hypotenuse and their sum would be the the net electric field

E1= (8.99 E 9)*(7.59 E -9)/ (0.045962)^2 = 32300.1

E2=(8.99 E 9)*(-1.09 E-8)/ (0.045962)^2 = -46386.2

E3= (8.99 E 9)*(1.15 E-8)/ (0.045962)^2 = 48939.5

E4= (8.99 E 9)*(-6.06 E-9)/ (0.045962)^2 = -25789

E net= 32300.1 + -46386.2 + 48939.5 + -25789 = 9064.45

I entered that in but my online homework is marking it as incorrect. Could someone please help. Thanks

2. Feb 4, 2012

### tiny-tim

Hi C6ZR1!
nooo

fields are vectors, so you can't just add their magnitudes (scalar addition) …

you must add them as vectors

3. Feb 4, 2012

### C6ZR1

so then for each point would I use coordinates and subtract that from the center, C (0.0325,0.0325) and go about getting each particle in vector form?

4. Feb 4, 2012

### tiny-tim

i'm not sure what you mean by that

do it the easy way …

each vector points along a diagonal, so add the opposite pairs first, and that will give you two perpendicular vectors to add

5. Feb 4, 2012

### C6ZR1

but how do we know its components if we are only given the points? Sorry if I'm over looking it. I literally just learned about vectors/cross and dot products this week in school.

6. Feb 4, 2012

### tiny-tim

technically, a vector isn't components, it's a line (with an arrow at the end of it)

so draw the lines (with arrows) …

each line will start at the centre, and point towards (or away from, if the force is repulsive) each of the four corners

(you could write out the components, but there's no point when the vectors themselves are so easy to draw)

7. Feb 4, 2012

### C6ZR1

ok, assuming its a positively charged point charge in the center then the arrows are are pointing towards, the center from Q1, and Q3. and towards Q3 and Q4

8. Feb 4, 2012

### tiny-tim

(you meant …)
yes

9. Feb 4, 2012

### C6ZR1

q1 and q3: 1.909E-8 C
q2+q4: -1.696 E-8

but I'm still having a little trouble trying to understand why we do this. Is it because those pairs are on a same line?

10. Feb 4, 2012

### tiny-tim

sorry, where do those figures come from?
yes, it's easy to add vectors that are in the same line

11. Feb 4, 2012

### C6ZR1

They were the charges of each particle in C. I dont think I'm understanding this too well. :/ What would I add?

12. Feb 4, 2012

### tiny-tim

ohhh

you should be adding the vectors (the fields)

13. Feb 4, 2012

### C6ZR1

Vector fields = Electric field produced by each particle?

14. Feb 5, 2012

### tiny-tim

(just got up :zzz: …)

yes

15. Feb 5, 2012

### C6ZR1

lol, well good morning.

So if the vector fields are electric fields produced by each particle then was my original attempt correct my finding the electric fields and adding them up?

16. Feb 5, 2012

### tiny-tim

you didn't add the fields, you only added (or subtracted) their magnitudes

17. Feb 5, 2012

### C6ZR1

ohhh, but isnt electric field E=(kq)/r^2?

18. Feb 5, 2012

### tiny-tim

no, its (kq)/r2 radially away from the source

it's a vector!!

potential is a scalar, field is a vector

19. Feb 5, 2012

### C6ZR1

so is K the positive test charge in the center of the field?

20. Feb 5, 2012

### tiny-tim

you mean the k in kq/r2?

no, that's just a constant