anonymous
  • anonymous
A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x = 0.60 m , y = 0. Part A)Calculate the magnitude of the net electric field at the origin due to these two point charges Part B)Calculate the direction of the net electric field at the origin due to these two point charges.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
experimentX
  • experimentX
can you calculate the electric field due to each charges separately?
anonymous
  • anonymous
Can I do it? No, I understand that's what I think I have to do, but I just don't know how :/
experimentX
  • experimentX
how do you calculate electric field due to a charge Q at origin at point (x,y)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
coulomb's law?
experimentX
  • experimentX
yes ... use coloumd law to calculate electric field.
experimentX
  • experimentX
first try my problem (it's easy) ... draw picture, then we will try yours.
anonymous
  • anonymous
alright
anonymous
  • anonymous
what is your problem?
experimentX
  • experimentX
a charge of magnitude Q is at origin, calculate electric field at (3,4)
anonymous
  • anonymous
Just to make sure is this the equation I have to use? \[F = k (|q_1q_2|)/r^2\]
experimentX
  • experimentX
yes ... E = F/q
anonymous
  • anonymous
how do I find r? that's one of the things that has been confusing me
experimentX
  • experimentX
r= sqrt(x^2+y^2)
anonymous
  • anonymous
oooh
anonymous
  • anonymous
would it be: \[F = ((9 x 10^9) (Q) )/ 5\]
anonymous
  • anonymous
that is 9 x 10^9
anonymous
  • anonymous
so, I'm guessing im wrong? :/
experimentX
  • experimentX
r^2
anonymous
  • anonymous
so would that be 25 instead of 5 then.
experimentX
  • experimentX
yes
experimentX
  • experimentX
is Electric Field scalar or vector?
anonymous
  • anonymous
vector
experimentX
  • experimentX
do you know that vector has components?
anonymous
  • anonymous
yea
experimentX
  • experimentX
how do you find the two components of this vector?
experimentX
  • experimentX
draw figure ...
anonymous
  • anonymous
|dw:1378573530590:dw|
experimentX
  • experimentX
|dw:1378573584100:dw| do you know how to find Ex and Ey?
anonymous
  • anonymous
Ex = 3cos(angle) Ey = 4sin(angle)
anonymous
  • anonymous
I think :/
experimentX
  • experimentX
No .. must be Ex = E cos(angle) Ey = E sin(angle) find angle and put the value of E from above that you calculated.
anonymous
  • anonymous
Oh okay, but to find the value of E don't I still need a value for Q, you never told me what the charge is equal to, you just said it was Q
experimentX
  • experimentX
put put it Q for a while ... it doesn't matter ... express it interms of Q if you like you can put Q=1C
anonymous
  • anonymous
alright
anonymous
  • anonymous
I got E = 3.6 x 10^8
experimentX
  • experimentX
seems okay http://www.wolframalpha.com/input/?i=9*10%5E9%2F25 find angle
anonymous
  • anonymous
But I'm still missing what Ex and Ey equal. Like I have two unknown parameters
anonymous
  • anonymous
like let's say Ex = Ecos(theta) I'm missing Ex and theta
experimentX
  • experimentX
angle ... calculate the angle
experimentX
  • experimentX
|dw:1378574318729:dw|
anonymous
  • anonymous
Theta = Cos^-1 (Ex / E ) ?
experimentX
  • experimentX
|dw:1378574369513:dw|
anonymous
  • anonymous
i got 90 degrees
experimentX
  • experimentX
|dw:1378574479299:dw|
anonymous
  • anonymous
okay 53.13?
experimentX
  • experimentX
use that ...
anonymous
  • anonymous
Ex = 1.98 x 10^8 Ey = 2.64 x 10^8
experimentX
  • experimentX
let me give you another problem.|dw:1378574722719:dw|
experimentX
  • experimentX
|dw:1378574881944:dw|
anonymous
  • anonymous
oh ok, just one thing i was still confused on earlier did you say that r^2 = (sqrt) x^2 + y^2 OR r = (sqrt)x^2 + y^2
experimentX
  • experimentX
\[r^2 = x^2 + y^2 \] pythagoras formula
anonymous
  • anonymous
it should be the second one
anonymous
  • anonymous
yea
anonymous
  • anonymous
im working on them atm, im almost done
experimentX
  • experimentX
did you calculate Ex and Ey for this?
anonymous
  • anonymous
working on it right now
anonymous
  • anonymous
r^2 = 20 (theta) = 63.435 E = 9 x 10^8 Ex = 4.02 x 10^8 Ey = 8.05 x 10^8 I hope it's right >.<
experimentX
  • experimentX
all righ t... good. |dw:1378575628864:dw| Now i put both of them here.
anonymous
  • anonymous
Would I just add my answers from the two previous questions?
experimentX
  • experimentX
no ... not exactly. you add Ex1 + Ex2 and Ey1 + Ey2
anonymous
  • anonymous
oh ok. but technically wouldnt my Ex2 and Ey2 be the same, since it's the same charge in the same positin. I would just have to calculate Ex1 and Ey1 from the origin using a 2C charge instead of the 1C charge before
experimentX
  • experimentX
No ..
anonymous
  • anonymous
oh
experimentX
  • experimentX
They are the components ... X components add up to X and Y to Y
experimentX
  • experimentX
finally find total Ex and total Ey
anonymous
  • anonymous
alright
anonymous
  • anonymous
I got answers but idk it if it's right, but I did try my best
anonymous
  • anonymous
Ex = 8.702 x 10^8 Ey = 1.4525 x 10^9
anonymous
  • anonymous
sorry that took long
experimentX
  • experimentX
|dw:1378576951064:dw| where is your grand Ex and Ey,, use pythagoras theorem to calculate the grand E
anonymous
  • anonymous
oops, forgot that one
anonymous
  • anonymous
Etotal = 1.6947 x 10^9
experimentX
  • experimentX
also find the direction.
anonymous
  • anonymous
If I understood the question correctly, Theta = 59.10 degrees?
experimentX
  • experimentX
maybe ... i didn't calculate. but you know how to do your problem right?
anonymous
  • anonymous
i can try. but it'd be nice if I had someone confirm it afterwards
anonymous
  • anonymous
my problem is similar to the 3rd example you just showed me right
experimentX
  • experimentX
sure ... you show me what you did step my setp, i'll confirm.
anonymous
  • anonymous
Alright, i'll start with a sketch
anonymous
  • anonymous
Oh just to make sure. My charges are in nC, do I convert to C? also my locations are in meters, do I convert to cm?
anonymous
  • anonymous
|dw:1378577908308:dw|
anonymous
  • anonymous
is the sketch correct?
anonymous
  • anonymous
you there?
experimentX
  • experimentX
sorry ... i was lost. you should have pinged.
experimentX
  • experimentX
|dw:1378578909800:dw|
anonymous
  • anonymous
i didnt know how to ping. but how come q2 is 10^-9 but q1 is 10^9
experimentX
  • experimentX
@sean1372 nC = nano Coloumb do you know how huge 1C charge is ?
anonymous
  • anonymous
wait what? im confused now
experimentX
  • experimentX
1nC = 1x10^-C Coulomb. 1 coloumb is a huge amount of charge.
anonymous
  • anonymous
ooh since q1 is a -4.00 nC the negatives cancel out so that's what makes it a -4 x 10^9 and since the other charge is +6 it becomes a 6 x 10^-9
experimentX
  • experimentX
No ...
anonymous
  • anonymous
so the charges are -4.00 x 10^-9 C and 6 x 10^-9 C?
anonymous
  • anonymous
ive spent nearly 3 hours on this problem now.....
experimentX
  • experimentX
yes
anonymous
  • anonymous
are you saying yes to the charges?
experimentX
  • experimentX
yes -4nC = -4/1000000000 6nC = 6/1000000000
anonymous
  • anonymous
okay can you check this so far. E1 = (kq1) / r^2 = (9 x 10^9 )(-4 x 10^-9) / 1 = -36 E2 = (kq2) / r^2 = (9 x 10^-9) (6 x 10^-9) / 0.36 = 150
experimentX
  • experimentX
seems okay ... calculate the components.
anonymous
  • anonymous
awesome, alright let me get on it :)
experimentX
  • experimentX
also careful about sign
anonymous
  • anonymous
on which one?
experimentX
  • experimentX
one is minus and another is +
experimentX
  • experimentX
instead of adding you will be subtracting.
anonymous
  • anonymous
oh yea. gotcha
anonymous
  • anonymous
I got: Ex1 = -21.60 Ey1 = -28.6 Ex2 = 150 Ey2 = 0 Ex = Ex1 + Ex2 = (-21.60) + 150 = 128.4 Ey = Ey1 + Ey2 = -28.8 + 0 = -28.8 E = (sqrt) Ex^2 + Ey^2 = (sqrt) (128.4)^2 + (-28.8)^2 = 131.59 N/C Direction: (phi) = cos^-1 (128.4 / 131.59) = 12.64 degrees
experimentX
  • experimentX
all right ... seems okay
anonymous
  • anonymous
yey I just put in my answer for part A (132 N/C) and it was right
anonymous
  • anonymous
part B was wrong though this is what it says: Calculate the direction of the net electric field at the origin due to these two point charges (Phi) = ______________ ∘ counterclockwise from +x-axis
experimentX
  • experimentX
(phi) = cos^-1 (-28.8/128.4)
anonymous
  • anonymous
it said that was wrong to
experimentX
  • experimentX
that can't be wrong ...just make wrt x axis.
anonymous
  • anonymous
does the "Counterclockwise from the +x direction" have to do with anything
experimentX
  • experimentX
if i were you i would do 2pi - tan^(-1)(..____)
experimentX
  • experimentX
how much degree will you rotate?
anonymous
  • anonymous
wait a minute... you told me to put cos^-1 (-28.8 / 128.4) I think you meant to put tan^-1(-28.8 / 128.4)
experimentX
  • experimentX
sorry ... i didn't see that ... i copied your text and edited.
anonymous
  • anonymous
it's fine. I got -12.64 instead now
experimentX
  • experimentX
if you want to find angle, cosine is a bad choice. cosine is +ve on first and 4the quadrant.
anonymous
  • anonymous
yea, but i dont think the answer would be -12.64 either. because i already tried +12.64 and usually if the answer is correct, but you have the wrong sign, it just tells me to check my signs. but it just told me that it was wrong
anonymous
  • anonymous
I mean I can still try it, but i hate to keep loosing point on it
experimentX
  • experimentX
|dw:1378581859758:dw|
anonymous
  • anonymous
i don't know what to do with that pic right now
experimentX
  • experimentX
Ex = Ex1 + Ex2 = (-21.60) + 150 = 128.4 Ey = Ey1 + Ey2 = -28.8 + 0 = -28.8|dw:1378582017447:dw|
experimentX
  • experimentX
you want answer is degree or radians?
anonymous
  • anonymous
degrees
experimentX
  • experimentX
90 + 12.64 http://www.wolframalpha.com/input/?i=arctan%2828.8%2F128.4%29
anonymous
  • anonymous
no, it didn't work either.
experimentX
  • experimentX
that's surprising
anonymous
  • anonymous
Calculate the direction of the net electric field at the origin due to these two point charges. ϕ = ∘ counterclockwise from +x-axis
anonymous
  • anonymous
i don't know what we're doing wrong. I didn't think the angle part would be more difficult then the electric field calculation
experimentX
  • experimentX
http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html
anonymous
  • anonymous
what do I do with that
experimentX
  • experimentX
try 167.36
anonymous
  • anonymous
oh dang, it worked! what did you do?
experimentX
  • experimentX
magic
experimentX
  • experimentX
|dw:1378582687436:dw|
anonymous
  • anonymous
oooh you did 180 - 12.64
experimentX
  • experimentX
|dw:1378582825574:dw| + means go away - means come towards rest is just diagram, vectors, coulomb and Pythagoras.
anonymous
  • anonymous
I can't thank you enough. I know we spent a lot of time on this but I actually learned how to do this sort of problem. I know I was a bit slow at the start, I had just woken up and wasn't fully focused yet. But again. thanks for the help man
experimentX
  • experimentX
yw
anonymous
  • anonymous
well I guess I'll see you around :)
experimentX
  • experimentX
sure
goformit100
  • goformit100
"Welcome to OpenStudy. I can guide regarding this useful site; ask your doubts from me, for it you can message me. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it."

Looking for something else?

Not the answer you are looking for? Search for more explanations.