Four point charges of equal magnitude Q = 55 nC are placed on the corners of a r

Question

Four point charges of equal magnitude Q = 55 nC are placed on the corners of a rectangle of sides D1 = 25 cm and D2 = 12 cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. Use a coordinate system fixed to the bottom left hand charge, with positive directions as shown in the figure.

B) Calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the triangle. Fx=?

C) Calculate the vertical component of the net force, in newtons, on the charge which lies at the lower left corner of the triangle Fy=?

E) Calculate the angle of the net force vector, measured counterclockwise from the positive horizontal axis. Enter an angle between -180° and 180°. theta=?

Explanation / Answer

b] using Colombs law,

Horizontal force = summation kQ1Q2/r^2

= 9e9*55e-9^2/0.25^2 +  9e9*55e-9^2/(0.25^2+0.12^2) *0.25/sqrt(0.25^2+0.12^2)

= 0.000755 N

c] Vertical force = summation kQ1Q2/r^2

= -9e9*55e-9^2/0.12^2 +  9e9*55e-9^2/(0.25^2+0.12^2) *0.12/sqrt(0.25^2+0.12^2)

= -0.001737 N

d] Net force = sqrt(0.000755^2 + 0.001737^2) = 0.001894 N

e] angle = arctan(-0.001737/0.000755) = -66.5 degree

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