Question

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?

Answer 1

counterclockwise from the positive x axis

Answer 2

plz tell me u can do this?

Answer 3

yeah let me try

Answer 4

i got 8 minutes to submit answer

Answer 5

i hate timed tests

Answer 6

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.

Answer 7

chris i don't have the time...i got 7 minutes

Answer 8

Enough time to think ...

Answer 9

afterwards i'll worry about how's and why's...right now i just need the correct answer

Answer 10

26.77 is net electric charge

Answer 11

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

Answer 12

Second hint: Use coulomb's law to calculate the contribution of each charge at the origin.

Answer 13

plz for the love of god just gimme the answer and i'll work through it after i submit it

Answer 14

plz

Answer 15

got it?

Answer 16

I'm not going to support you to cheat, sorry.

Answer 17

then leave

Answer 18

ashe no i don't

Answer 19

are there options

Answer 20

Damn, just add these numbers.

Answer 21

what numbers?

Answer 22

i got 2 minutes left?

Answer 23

u wnana throw an airball up?

Answer 24

net electric field= \[\sqrt{{48.0355}^2+{4.028}^2}\]=48.204 Volt/meter in clockwise direction from =ve x-axis

Answer 25

it'd be +ve x -axis

Answer 26

i submitted 128...i think it's 132 though

Answer 27

thx for your help sir

Answer 28

but from where you got 128

Answer 29

i appreciate the effort

Answer 30

i added the two numbers u had earlier....as per what chris said

Answer 31

then just subtracted taht from 180

Answer 32

to get the angle counterclockwise

Answer 33

angle is -175 degrees

Answer 34

that was magnitude

Answer 35

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).

Answer 36

yea thx....coulda used that 15 minutes ago

Answer 37

good lookin out

Answer 38

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.

Answer 39

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