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Field due to Two Point Charges (will rate for a life saver !)

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Scholar
Karma Points: 225
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Date Posted: 6/18/2008 9:03:05 AM  Status: Closed
Field due to Two Point Charges (will rate for a life saver !)
Course Textbook Chapter Problem
Calculus Based Physics University Physics with Modern Physics (12th) by Young, Freedman 21 N/A
Question Details:
Two point charges are placed on the x axis.  View Figure  The first charge, q_1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q_2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis.
 
Part A
 
part B
Part A
Find the electric field at the origin, point O.
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.
E_{{\rm O}x}, E_{{\rm O}y}  =   \rm{N/C}
 
 
 
Part B
Now, assume that charge q_2 is negative; q_2 = -6 \; \rm {nC}.  View Figure  What is the net electric field at the origin, point O?
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.
  E_{{\rm O}x}, E_{{\rm O}y}  =
  \rm{N/C}
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Oracle
Karma Points: 14,566
Date Posted: 7/4/2008 2:29:47 AM  Status: Live
Asker's Rating: Lifesaver   
Response:
A)
Eox = -kq1/162 + kq2/92 = -9*109*8*10-9/162 + 9*109*6*10-9/92 = 0.385 N/C
Eoy = 0
B)
Eox = -kq1/162 + kq2/92 = -9*109*8*10-9/162 - 9*109*6*10-9/92 = -0.948 N/C
Eoy = 0

sadgirl's Comment:
thanks for the perfect answers




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