A +14 nC charge is located at (x,y)=(0cm, 12 cm ) and a -3.0 nC charge is locate
ID: 250669 • Letter: A
Question
A +14 nC charge is located at (x,y)=(0cm, 12 cm ) and a -3.0 nC charge is located (x,y)=( 4.0 cm ,0cm).
Enter your answers numerically separated by a comma. Express your answer using two significant figures.
Problem 20.59 Part A A +14 nC charge is located at (x,y) = (0 cm, 12cm ) and a-3.0 nC charge is located (x , y) ( 4.0 cm, 0 cm). Where would a -8.0 nC charge need to be located in order that the electric field at the origin be zero? Enter your answers numerically separated by a comma. Express your answer using two significant figures. z, y= cm Submit My Answers Give UpExplanation / Answer
Calculate the E field at 0,0 and then place the -8nC charge to exactly oppose it in magnitude and direction by adjusting it's distance to be equal in magnitude and it's location to be exactly 180° different in direction.
The directions of the fields at the origin due to the 14nC and -3nC charges are:
The +14nC field points down at 270° . The -3nC field points at 0° because + charge fields point OUT and - charge fields point IN
We know E=kq/r²
Plug in values: .
Ex = k*3e-9/0.04² = 16875 N/C at 0°
Ey = k*14e-9/0.12² = 8750 N/C at 270°
The fields are vectors and perpendicular so use Pythagoras theorem:
E = Ex² + Ey² ) = 19008.6 at 332.6° = arctan(Ey/Ex)
So the location of the -8nC will be 332.6°-180° = 152.6° because the E field from the -8nC charge points INTO the -8nC charge in exactly the opposite direction to the field at the origin.
The E field from the -8nC charge must cancel the field at the origin so
k*8e-9/r² = 19008.6
thus distance from the origin will be r² = k*8e-9/19008.6 => r = 6.15 cm
y = 6.15*sin152.6 = 2.83 cm
x = 6.15*cos152.6 = -5.46 cm
x,y= (-5.46 cm, 2.83 cm)
Hope this helps!!!
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