Question

A charge of -3.05 nC is placed at the origin of an xy-coordinate system, and a charge of 2.45 nC is placed on the y axis at y = 3.80 cm .

part A) If a third charge, of 5.00 nC , is now placed at the point x = 3.05 cm , y = 3.80 cm find the x and y components of the total force exerted on this charge by the other two charges.

part B) Find the magnitude of this force.

part C) Find the direction of this force.

Answer #1

force between q1 and q3

F13 = k q3 q1 / r13^2 = 9* 10^9* 3.05* 5* 10^-18 / (0.0305^2 + 0.038^2)

F13 = 5.7808 * 10^-5 N

angle formed by F13 with horizontal

x = arctan ( 3.8 / 3.05) = 51.248

force between q2 and q3

F23 = 9* 10^9* 2.45* 5* 10^-18 / 0.038^2

F23 = 7.635* 10^-5 N

======

a)

Net horizontal force

Fx = F23 - F13 cos x = - 4.017 * 10^-5 N

net vertical force

Fy = F13 sin 51.248 = - 4.508 * 10^-5 N

=====

b)

magnitude

F^2 = Fx^2 + Fy^2

F = 6.038 * 10^-5 N

======

c)

direction

phi = arctan ( Fy/FX) = 48.3 BELOW +X AXIS

=======

Comment before rate in case any doubt, will reply for sure.. goodluck

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