Question

Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q 5.28 nC at the origin. (Let r,2 = 0.325 m.) 6.00 nC 12 0.100 m -3.00 nC magnitude N oCounterclockwise from the -x (negative x) axis direction

Answer 1

6.00 nc 9 5.28 nc 0.325 m + + Fx 0.100 m -3.00 nC Fy qy

Use formula $F_{e}=\frac{kq_{1}q_{2}}{d^{2}}$

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Consider the horizontal force

$F_{x}=\frac{kq_{x}q}{x^{2}}$

$F_{x}=\frac{9*10^{9}*6*10^{-9}C*5.28*10^{-9}C}{(0.325m)^{2}}$

$F_{x}=\frac{9*6*5.28*10^{-9}}{0.325^{2}}$

$F_{x}=2.6994*10^{-6}N$

F_x is acting along -ve x-direction, so put a negative sign

${\color{Red} F_{x}=-2.6994*10^{-6}N}$

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Consider vertical forces

$F_{y}=\frac{kq_{y}q}{y^{2}}$

$F_{y}=\frac{9*10^{9}*3*10^{-9}C*5.28*10^{-9}C}{(0.1m)^{2}}$

$F_{y}=\frac{9*3*5.28*10^{-9}}{0.1^{2}}$

$F_{y}=1.4256*10^{-5}N$

F_y is acting along -ve y-direction, so put a negative sign

${\color{Red} F_{y}=-1.4256*10^{-5}N}$

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Magnitude, $F=\sqrt{F_{x}^{2}+F_{y}^{2}}$

$F=\sqrt{(-2.6994*10^{-6}N)^{2}+(-1.4256*10^{-5}N)^{2}}$

ANSWER: ${\color{Red} F=1.4509*10^{-5}N}$

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Direction

$\theta =tan^{-1}(\frac{F_{y}}{F_{x}})$

$\theta =tan^{-1}(\frac{-1.4256*10^{-5}N}{-2.6994*10^{-6}N})$

$\theta =tan^{-1}(\frac{14.256}{2.6994})$

$\theta =79.28^{\circ}$ third quadrant

ANSWER: ${\color{Red} \theta =79.28^{\circ}}$ counterclockwise from the -ve x-axis

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