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E field

by sasuke07
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sasuke07
#1
Mar8-12, 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
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tiny-tim
#2
Mar8-12, 06:33 AM
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hi sasuke07!

(try using the X2 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?
sasuke07
#3
Mar8-12, 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.

tiny-tim
#4
Mar8-12, 06:56 AM
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E field

Quote Quote by sasuke07 View Post
WOuldn't the negative charge have the same magnitude of the positive charge.
yes
What do you mean by the x and y components.
draw an arrow at the origin showing the direction of the field from the positive charge

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 )
sasuke07
#5
Mar8-12, 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.
tiny-tim
#6
Mar8-12, 07:09 AM
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no, you must use cos and sin, of the angle the arrow makes
sasuke07
#7
Mar8-12, 07:11 AM
P: 54
so would it be Kq/r^2sintheta
and Kq/r^2Costheta. Where theta would be 90 degrees?
sasuke07
#8
Mar8-12, 07:14 AM
P: 54
or would theta be 45
tiny-tim
#9
Mar8-12, 07:18 AM
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θ is the angle of the line from the charge to the origin
sasuke07
#10
Mar8-12, 07:19 AM
P: 54
awesome so 45 degrees.
sasuke07
#11
Mar8-12, 07:19 AM
P: 54
So were the equations correct though?
tiny-tim
#12
Mar8-12, 07:30 AM
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looks ok
so what is the total field from both the positive and the negative charge?
sasuke07
#13
Mar8-12, 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
tiny-tim
#14
Mar8-12, 07:48 AM
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Quote Quote by sasuke07 View Post
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
your answer looks right to me

(are you sure it's 5 cm that they're asking about?)
sasuke07
#15
Mar8-12, 07:52 AM
P: 54
i think its 5cm because the length of the x and y components is 5cm.


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