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E fieldby sasuke07
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#1
Mar812, 05:02 AM

P: 54

1. The problem statement, all variables and given/known data
What is the magnitude of the collective contribution to the E field at the origin from both the + and  charges located 5 cm above the horizontal axis? 2. Relevant equations kQ/d^2 3. The attempt at a solution Im just having a problem setting up this problem. Could someone help me on getting started 


#2
Mar812, 06:33 AM

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hi sasuke07!
(try using the X^{2} button just above the Reply box ) first, find the magnitude of each contribution, from that, find the x and y components … what do you get? 


#3
Mar812, 06:47 AM

P: 54

I already figured out the magnitued of the the positive charge 5 cm above the axis and got 5.4X10^4. What do you mean by the x and y components. WOuldn't the negative charge have the same magnitude of the positive charge.



#4
Mar812, 06:56 AM

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E field
then find the x and y components of that arrow (and finally you'll add them to the x and y components of the arrow from the negative charge ) 


#5
Mar812, 07:02 AM

P: 54

so to figure out the x component wouldn't it be Kq/d^2 where k=9X10^9, q is 30X10^9= charge and d would be .05m^2. Or would i have to use cosine and sine to figure out the x and y components and if so could you show me how to set it up.



#6
Mar812, 07:09 AM

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no, you must use cos and sin, of the angle the arrow makes



#7
Mar812, 07:11 AM

P: 54

so would it be Kq/r^2sintheta
and Kq/r^2Costheta. Where theta would be 90 degrees? 


#8
Mar812, 07:14 AM

P: 54

or would theta be 45



#9
Mar812, 07:18 AM

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θ is the angle of the line from the charge to the origin



#10
Mar812, 07:19 AM

P: 54

awesome so 45 degrees.



#11
Mar812, 07:19 AM

P: 54

So were the equations correct though?



#12
Mar812, 07:30 AM

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looks ok
so what is the total field from both the positive and the negative charge? 


#13
Mar812, 07:32 AM

P: 54

i got 7.6X10^4 by doing KQ/r^2sin45 and the answer is the same for cos45. BUt i checked the answer and its supposed to be 7.1X10^4. Any suggestions



#14
Mar812, 07:48 AM

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(are you sure it's 5 cm that they're asking about?) 


#15
Mar812, 07:52 AM

P: 54

i think its 5cm because the length of the x and y components is 5cm.



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