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.

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- experimentX

can you calculate the electric field due to each charges separately?

- 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

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

coulomb's law?

- experimentX

yes ... use coloumd law to calculate electric field.

- experimentX

first try my problem (it's easy) ... draw picture,
then we will try yours.

- anonymous

alright

- anonymous

what is your problem?

- experimentX

a charge of magnitude Q is at origin,
calculate electric field at (3,4)

- anonymous

Just to make sure is this the equation I have to use?
\[F = k (|q_1q_2|)/r^2\]

- experimentX

yes ... E = F/q

- anonymous

how do I find r? that's one of the things that has been confusing me

- experimentX

r= sqrt(x^2+y^2)

- anonymous

oooh

- anonymous

would it be:
\[F = ((9 x 10^9) (Q) )/ 5\]

- anonymous

that is 9 x 10^9

- anonymous

so, I'm guessing im wrong? :/

- experimentX

r^2

- anonymous

so would that be 25 instead of 5 then.

- experimentX

yes

- experimentX

is Electric Field scalar or vector?

- anonymous

vector

- experimentX

do you know that vector has components?

- anonymous

yea

- experimentX

how do you find the two components of this vector?

- experimentX

draw figure ...

- anonymous

|dw:1378573530590:dw|

- experimentX

|dw:1378573584100:dw|
do you know how to find Ex and Ey?

- anonymous

Ex = 3cos(angle)
Ey = 4sin(angle)

- anonymous

I think :/

- 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

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

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

alright

- anonymous

I got E = 3.6 x 10^8

- experimentX

seems okay
http://www.wolframalpha.com/input/?i=9*10%5E9%2F25
find angle

- anonymous

But I'm still missing what Ex and Ey equal. Like I have two unknown parameters

- anonymous

like let's say Ex = Ecos(theta)
I'm missing Ex and theta

- experimentX

angle ... calculate the angle

- experimentX

|dw:1378574318729:dw|

- anonymous

Theta = Cos^-1 (Ex / E ) ?

- experimentX

|dw:1378574369513:dw|

- anonymous

i got 90 degrees

- experimentX

|dw:1378574479299:dw|

- anonymous

okay 53.13?

- experimentX

use that ...

- anonymous

Ex = 1.98 x 10^8
Ey = 2.64 x 10^8

- experimentX

let me give you another problem.|dw:1378574722719:dw|

- experimentX

|dw:1378574881944:dw|

- 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

\[r^2 = x^2 + y^2 \]
pythagoras formula

- anonymous

it should be the second one

- anonymous

yea

- anonymous

im working on them atm, im almost done

- experimentX

did you calculate Ex and Ey for this?

- anonymous

working on it right now

- 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

all righ t... good.
|dw:1378575628864:dw|
Now i put both of them here.

- anonymous

Would I just add my answers from the two previous questions?

- experimentX

no ... not exactly.
you add Ex1 + Ex2
and Ey1 + Ey2

- 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

No ..

- anonymous

oh

- experimentX

They are the components ...
X components add up to X
and Y to Y

- experimentX

finally find total Ex and total Ey

- anonymous

alright

- anonymous

I got answers but idk it if it's right, but I did try my best

- anonymous

Ex = 8.702 x 10^8
Ey = 1.4525 x 10^9

- anonymous

sorry that took long

- experimentX

|dw:1378576951064:dw|
where is your grand Ex and Ey,, use pythagoras theorem to calculate the grand E

- anonymous

oops, forgot that one

- anonymous

Etotal = 1.6947 x 10^9

- experimentX

also find the direction.

- anonymous

If I understood the question correctly,
Theta = 59.10 degrees?

- experimentX

maybe ... i didn't calculate.
but you know how to do your problem right?

- anonymous

i can try. but it'd be nice if I had someone confirm it afterwards

- anonymous

my problem is similar to the 3rd example you just showed me right

- experimentX

sure ... you show me what you did step my setp, i'll confirm.

- anonymous

Alright, i'll start with a sketch

- 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

|dw:1378577908308:dw|

- anonymous

is the sketch correct?

- anonymous

you there?

- experimentX

sorry ... i was lost. you should have pinged.

- experimentX

|dw:1378578909800:dw|

- anonymous

i didnt know how to ping. but how come q2 is 10^-9 but q1 is 10^9

- experimentX

@sean1372 nC = nano Coloumb
do you know how huge 1C charge is ?

- anonymous

wait what? im confused now

- experimentX

1nC = 1x10^-C Coulomb.
1 coloumb is a huge amount of charge.

- 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

No ...

- anonymous

so the charges are -4.00 x 10^-9 C and 6 x 10^-9 C?

- anonymous

ive spent nearly 3 hours on this problem now.....

- experimentX

yes

- anonymous

are you saying yes to the charges?

- experimentX

yes
-4nC = -4/1000000000
6nC = 6/1000000000

- 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

seems okay ... calculate the components.

- anonymous

awesome, alright let me get on it :)

- experimentX

also careful about sign

- anonymous

on which one?

- experimentX

one is minus and another is +

- experimentX

instead of adding you will be subtracting.

- anonymous

oh yea. gotcha

- 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

all right ... seems okay

- anonymous

yey I just put in my answer for part A (132 N/C) and it was right

- 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) = ______________ âˆ˜ counterclockwise from +x-axis

- experimentX

(phi) = cos^-1 (-28.8/128.4)

- anonymous

it said that was wrong to

- experimentX

that can't be wrong ...just make wrt x axis.

- anonymous

does the "Counterclockwise from the +x direction" have to do with anything

- experimentX

if i were you i would do 2pi - tan^(-1)(..____)

- experimentX

how much degree will you rotate?

- 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

sorry ... i didn't see that ... i copied your text and edited.

- anonymous

it's fine. I got -12.64 instead now

- experimentX

if you want to find angle, cosine is a bad choice. cosine is +ve on first and
4the quadrant.

- 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

I mean I can still try it, but i hate to keep loosing point on it

- experimentX

|dw:1378581859758:dw|

- anonymous

i don't know what to do with that pic right now

- experimentX

Ex = Ex1 + Ex2
= (-21.60) + 150
= 128.4
Ey = Ey1 + Ey2
= -28.8 + 0
= -28.8|dw:1378582017447:dw|

- experimentX

you want answer is degree or radians?

- anonymous

degrees

- experimentX

90 + 12.64
http://www.wolframalpha.com/input/?i=arctan%2828.8%2F128.4%29

- anonymous

no, it didn't work either.

- experimentX

that's surprising

- anonymous

Calculate the direction of the net electric field at the origin due to these two point charges.
Ï• = âˆ˜ counterclockwise from +x-axis

- 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

http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html

- anonymous

what do I do with that

- experimentX

try 167.36

- anonymous

oh dang, it worked! what did you do?

- experimentX

magic

- experimentX

|dw:1378582687436:dw|

- anonymous

oooh you did 180 - 12.64

- experimentX

|dw:1378582825574:dw|
+ means go away
- means come towards
rest is just diagram, vectors, coulomb and Pythagoras.

- 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

yw

- anonymous

well I guess I'll see you around :)

- experimentX

sure

- 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.