anonymous
  • anonymous
just need this answer real quick...timed test..got 13 minutes left: A +48.26 nC point charge is placed on the x axis at 3.007 m, and a -21.49 nC point charge is placed on the y axis at y= -6.779 m. What is the direction of the electric field at the origin?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
counterclockwise from the positive x axis
anonymous
  • anonymous
plz tell me u can do this?
ash2326
  • ash2326
yeah let me try

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More answers

ash2326
  • ash2326
|dw:1328301704314:dw|
anonymous
  • anonymous
i got 8 minutes to submit answer
anonymous
  • anonymous
i hate timed tests
anonymous
  • anonymous
Just think about it yourself. First hint: You can regard the resulting electrical field as the sum of the two electrical fields of each charge.
anonymous
  • anonymous
chris i don't have the time...i got 7 minutes
anonymous
  • anonymous
Enough time to think ...
anonymous
  • anonymous
afterwards i'll worry about how's and why's...right now i just need the correct answer
anonymous
  • anonymous
26.77 is net electric charge
ash2326
  • ash2326
Electric field at origin due to positive charge=9*10^9*48.26*10^-9/(3.007)^2( -i) = 48.0355 (-i) (toward left ) Electric field at origin due to negative charge=9*10^9*(-21.49)*10^-9/(6.779)^2 (j) =4.208(-j) towards downward direction I think clockwise from x direction
anonymous
  • anonymous
Second hint: Use coulomb's law to calculate the contribution of each charge at the origin.
anonymous
  • anonymous
plz for the love of god just gimme the answer and i'll work through it after i submit it
anonymous
  • anonymous
plz
ash2326
  • ash2326
got it?
anonymous
  • anonymous
I'm not going to support you to cheat, sorry.
anonymous
  • anonymous
then leave
anonymous
  • anonymous
ashe no i don't
ash2326
  • ash2326
are there options
anonymous
  • anonymous
Damn, just add these numbers.
anonymous
  • anonymous
what numbers?
anonymous
  • anonymous
i got 2 minutes left?
anonymous
  • anonymous
u wnana throw an airball up?
ash2326
  • ash2326
net electric field= \[\sqrt{{48.0355}^2+{4.028}^2}\]=48.204 Volt/meter in clockwise direction from =ve x-axis
ash2326
  • ash2326
it'd be +ve x -axis
anonymous
  • anonymous
i submitted 128...i think it's 132 though
anonymous
  • anonymous
thx for your help sir
ash2326
  • ash2326
but from where you got 128
anonymous
  • anonymous
i appreciate the effort
anonymous
  • anonymous
i added the two numbers u had earlier....as per what chris said
anonymous
  • anonymous
then just subtracted taht from 180
anonymous
  • anonymous
to get the angle counterclockwise
ash2326
  • ash2326
angle is -175 degrees
ash2326
  • ash2326
that was magnitude
anonymous
  • anonymous
If ash's numbers are correct it's just \[\vec E = ({-48.0},{-4.2} )^T \frac{\textrm{N}}{\textrm{C}}\], i.e. 180.5° (counter-clockwise).
anonymous
  • anonymous
yea thx....coulda used that 15 minutes ago
anonymous
  • anonymous
good lookin out
Xishem
  • Xishem
In the future you might consider: a.) Not wait until the last minute to do a test like this. b.) Being respectful to those who helped you. They aren't required to do so.
anonymous
  • anonymous
to xishem....i did not wait until the last minute to do this test. It's a timed 20 minute 1 question test. I'm very respectful of those who help me. I help many many people on here in the math section. I was very respectful and appreciative of ash for taking the time to help me. Chris was not helping, therefore at the time, was not appreciative

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