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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0counterclockwise from the positive x axis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0plz tell me u can do this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got 8 minutes to submit answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just 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
 4 years ago
Best ResponseYou've already chosen the best response.0chris i don't have the time...i got 7 minutes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Enough time to think ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0afterwards i'll worry about how's and why's...right now i just need the correct answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.026.77 is net electric charge

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1Electric 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
 4 years ago
Best ResponseYou've already chosen the best response.0Second hint: Use coulomb's law to calculate the contribution of each charge at the origin.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0plz for the love of god just gimme the answer and i'll work through it after i submit it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not going to support you to cheat, sorry.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Damn, just add these numbers.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got 2 minutes left?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u wnana throw an airball up?

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1net electric field= \[\sqrt{{48.0355}^2+{4.028}^2}\]=48.204 Volt/meter in clockwise direction from =ve xaxis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i submitted 128...i think it's 132 though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thx for your help sir

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1but from where you got 128

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i appreciate the effort

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i added the two numbers u had earlier....as per what chris said

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then just subtracted taht from 180

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to get the angle counterclockwise

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If ash's numbers are correct it's just \[\vec E = ({48.0},{4.2} )^T \frac{\textrm{N}}{\textrm{C}}\], i.e. 180.5° (counterclockwise).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea thx....coulda used that 15 minutes ago

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In 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
 4 years ago
Best ResponseYou've already chosen the best response.0to 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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