Four Equidistant Particles in electric fields

[C6ZR1]

Homework Statement

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

Homework Equations

E=kq/r^2

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

 

[tiny-tim]

Hi C6ZR1!

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

nooo …

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

you must add them as vectors ​

 

[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?

 

[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

 

[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.

 

[tiny-tim]

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.

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)

 

[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

 

[tiny-tim]

(you meant …)

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 Q2 and Q4

yes

and now add each pair

 

[C6ZR1]

so adding
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?

 

[tiny-tim]

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

sorry, where do those figures come from?

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?

yes, it's easy to add vectors that are in the same line

 

[C6ZR1]

They were the charges of each particle in C. I don't think I'm understanding this too well. :/ What would I add?

 

[tiny-tim]

ohhh

you should be adding the vectors (the fields)

 

[C6ZR1]

Vector fields = Electric field produced by each particle?

 

[tiny-tim]

(just got up :zzz: …)

yes

 

[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?

 

[tiny-tim]

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

 

[C6ZR1]

ohhh, but isn't electric field E=(kq)/r^2?

 

[tiny-tim]

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

it's a vector!​

potential is a scalar, field is a vector

 

[C6ZR1]

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

 

[tiny-tim]

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

you mean the k in kq/r²?

no, that's just a constant

 

[C6ZR1]

omg, not k, I meant q. I have no idea why I typed k.

 

[tiny-tim]

omg, not k, I meant q. I have no idea why I typed k.

ohh!

no, the positive test charge is the Q in F = kqQ/r² …

the field E is the force-per-charge (force-per-test-charge, if you prefer),

so we divide F by Q to get E = force/charge = F/Q = kq/r²

 

[C6ZR1]

that makes sense, but if we have a constant Q in this case, the positive test charge, as well as equidistant particles from that test charge then wouldn't E be the same for each particle? ie. E1= 8.99E9*1.602E-19/ (0.045962)^2

Im kinda confused on how they want use to find the magnitude of the net electro static force in this case

 

[tiny-tim]

that makes sense, but if we have a constant Q in this case, the positive test charge, as well as equidistant particles from that test charge then wouldn't E be the same for each particle?

not following you

the actual charges are all different

 

[C6ZR1]

ohh wait, is it F=kq1q2/r^2 where q1 is one of the particles and q2 is the fixed positive test charge?

 

[tiny-tim]

yup!

 

[C6ZR1]

ohhhhhhhhhh, then in that case would I Just find F for each particle then add them up to get the sum?

 

[tiny-tim]

ohhhhhhhhhh, then in that case would I Just find F for each particle then add them up to get the sum?

yes …

provided you add them as vectors​

 

[C6ZR1]

gotcha, I'll do this when I have break between classes.

 

[C6ZR1]

ok so I found that with the formula F=kq1q2/r^2 in the diagonal direction where the charges are being repulsed that F= 6.50731 E-15 and diagonal direction where charge are drawn inward F=-37.1554, if my calculations are correct, would I just take the dot product and multiply them together to get the magnitude?

 

[tiny-tim]

ok so I found that with the formula F=kq1q2/r^2 in the diagonal direction where the charges are being repulsed that F= 6.50731 E-15 and diagonal direction where charge are drawn inward F=-37.1554, if my calculations are correct, would I just take the dot product and multiply them together to get the magnitude?

Sorry, C6ZR1, but you seem clueless as to what a vector is.

You don't add two perpendicular vectors by taking their dot product.

Let's try this …

If I add a vector of 3 in the x direction to 4 in the y direction, what do i get?​

 

[C6ZR1]

well that would be 3i+4j, correct?

Would I do the cross product?

 

[SammyS]

well that would be 3i+4j, correct?

That's correct for adding the vectors tiny-tim asked about.

Would I do the cross product?

tiny-tim is telling you to add the Electric Field due to each charge, adding them as vectors. There is NO product involved in that operation. It's adding, not multiplying.

 

[C6ZR1]

ok I think I'm completely lost.

I find the electric field from each charge:

E1= (8.99E9*1.602E-19*7.59E-9) /(0.065)^2= 2.58724 E-15 C

E2= (8.99E9*1.602E-19*-1.09E-8) /(0.065)^2= -3.71554 E-15 C

E3= (8.99E9*1.602E-19*1.15E8) /(0.065)^2= 39.2007 C

E4= (8.99E9*1.602E-19*-6.06E-9) /(0.065)^2= -2.0657E-15 C


Would I resolve them into their components? ie. E1= cos(45)2.58724 E-15 C+sin(45)2.58724 E-15 C etc?

Sorry for the trouble, I don't know why I'm not comprehending this

 

[tiny-tim]

Hi C6ZR1!

(just got up :zzz: …)

well that would be 3i+4j, correct?

yes

Would I do the cross product?

why?

as SammyS says, 3i+4j is the answer

when you add 3i to 4j (as vectors), the answer is 3i+4j

(obviously, it's more complicated if the two vectors aren't perpendicular )

(alternatively, you can say that it's 5 in the direction whose tan is 4/3)