Question

41 92 11.[1pt] Three point charges are located in an equilateral triangle with side 0.230 cm, where q1 the x component, then the y component 990 μC-2 +640 μ C, and 93 +2.50 nC What is the electric field at the location of q3? Enter Answer 1 of 2: Submit All Answers Last Answer 1: -2974 x 1043 N/C Last Answer 2:-155.2 x 10A4 N/C Answer 2 of 2: Incorrect, tries 2/15. 12. [1pt What is the force on q3? Enter the x component, then the y component Answer 1 of 2: Answer 2 of 2: Submit All Answers Last Answer 1: -6.54 N Last Answer 2:-52.76 N

Answer 1

93 93 60 E1 E1 E2 E2 60 de E2sin60 1in60 E2cos60 91 92

Electric field ,$E=\frac{kq}{r^{2}}$

$E_{1}=\frac{kq_{1}}{r^{2}}$

$E_{1}=\frac{9*10^{9}*9.9*10^{-6}C}{(2.3*10^{-3}m)^{2}}$

${\color{Red} E_{1}=1.684*10^{10}N/C}$

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$E_{2}=\frac{kq_{2}}{r^{2}}$

$E_{2}=\frac{9*10^{9}*6.4*10^{-6}C}{(2.3*10^{-3}m)^{2}}$

${\color{Red} E_{2}=1.089*10^{10}N/C}$

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$E_{x}=-E_{1}cos60-E_{2}cos60$

$E_{x}=-1.684*10^{10}cos60-1.089*10^{10}*cos60$

ANSWER: ${\color{Red} E_{x}=-1.3865*10^{10}N/C}$

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$E_{y}=E_{2}sin60-E_{1}sin60$

$E_{y}=1.089*10^{10}sin60-1.684*10^{10}sin60$

ANSWER: ${\color{Red} E_{y}=-5.153*10^{9}N/C}$

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Force, $F=Eq$

$F_{x}=E_{x}*q_{3}$

$F_{x}=-1.3865*10^{10}N/C*2.5*10^{-9}C$

ANSWER: ${\color{Red} F_{x}=-34.663N}$

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$F_{y}=E_{y}*q_{3}$

$F_{y}=-5.153*10^{9}N/C*2.5*10^{-9}C$

ANSWER: ${\color{Red} F_{y}=-12.883N}$

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