College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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case 3 is the first picture

O Case 4
This case is identical to Case 3 with one difference; qc = 11 µC.
A.) Without doing any calculations, determine the
magnitude & components of the electric force that C
exerts on E. Let up & right be positive.
FCE =
FCE,X
FCE,Y
%3D
B.) Without doing any calculations, determine the
magnitude & components of the electric force that D
exerts on E. Let up & right be positive.
FDE =
FDE,X
FDE,Y
D
E
C.) Determine the components, magnitude, & direction
of the net electric force on E because of C & D (you
will need to do calculations for this part).
Fnet,x
Fnet,y
Fnet =
direction of Fnet
° relative to the VERTICAL
%3D
in quadrant ( 2
expand button
Transcribed Image Text:O Case 4 This case is identical to Case 3 with one difference; qc = 11 µC. A.) Without doing any calculations, determine the magnitude & components of the electric force that C exerts on E. Let up & right be positive. FCE = FCE,X FCE,Y %3D B.) Without doing any calculations, determine the magnitude & components of the electric force that D exerts on E. Let up & right be positive. FDE = FDE,X FDE,Y D E C.) Determine the components, magnitude, & direction of the net electric force on E because of C & D (you will need to do calculations for this part). Fnet,x Fnet,y Fnet = direction of Fnet ° relative to the VERTICAL %3D in quadrant ( 2
As shown below, particle C is located at (0 cm, 19 cm), particle D is at the origin, & particle E is at (19 cm, 0 cm). Additionally, qc =
-11 µC, qp = 9 µC, & qɛ = -10 µC.
A.) Determine the magnitude & components of the
electric force that C exerts on E. Let up & right be
positive.
FCE = 1.37119e1 N
FCE,X
FCE,Y
9.695 N
= -9.695 N
B.) Determine the magnitude & components of the
electric force that D exerts on E. Let up & right be
positive.
FDE = 22.43 N
FDE,X
= -22.43 N
D
FDE,y
= ON
C.) Determine the components, magnitude, & direction
of the net electric force on E because of C & D.
Fnet,x
Fnet,y
Fnet =
direction of Fnet
HORIZONTAL in quadrant ( 3
%3D
° relative to the
%3D
expand button
Transcribed Image Text:As shown below, particle C is located at (0 cm, 19 cm), particle D is at the origin, & particle E is at (19 cm, 0 cm). Additionally, qc = -11 µC, qp = 9 µC, & qɛ = -10 µC. A.) Determine the magnitude & components of the electric force that C exerts on E. Let up & right be positive. FCE = 1.37119e1 N FCE,X FCE,Y 9.695 N = -9.695 N B.) Determine the magnitude & components of the electric force that D exerts on E. Let up & right be positive. FDE = 22.43 N FDE,X = -22.43 N D FDE,y = ON C.) Determine the components, magnitude, & direction of the net electric force on E because of C & D. Fnet,x Fnet,y Fnet = direction of Fnet HORIZONTAL in quadrant ( 3 %3D ° relative to the %3D
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