Question

The point charges in the figure have the following values: q₁ = +2.6·µC, q₂ = +5.5·µC, q₃ = -0.89·µC. (Let the +x axis point to the right.)

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(a) Given that the distance d in the figure is 4.35·cm, find the magnitude and direction of the net electrostatic force exerted on the point charge q₁.

(b) Find the magnitude and direction of the force in part (a) if the distance d is doubled.

(Give the direction in degrees counterclockwise from the +x axis)

Answer 1

-- -- 60° EK VE According to Coulomb's law F, = 9192 - 98 199 F2 4116 d2 2.6 x 10-6 x 5.5 x 10-6 4.35 x 10-4 -= 68 N 1 9193 - F341€ do 2.6 x 10-6 x 0.89 x 10-6 X 10' x =11 N 7.55 X 10-4 The angle between the forces is 120°, therefore the resultant of the force can be written as F = P3 + F3 + 2F,F2 cos 120° = 63.23 N

The direction of the force can be determine as follow, F, sino tana = F. + F, Cose 68 x sin 120° tan a = 1 11+ 68 x cos 120° 68 x 0.866 -1 = 1.3086 11 +68 X (-0.5) a = tan-21.3086 = 52.61 The direction of the force is -52.61° from the +x axis. Now of the distance of the charges is doubled then the force reduced by a factor of 4 because the force is inversely proportional to square of distance between the charges and there is no change in the direction of the force.

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