Question

A charge 4.98 nC is placed at the origin of an xy-coordinate system, and a charge -2.05 nC is placed on the positive x-axis at x = 3.97 cm . A third particle, of charge 6.04 nC is now placed at the point x = 3.97 cm , y = 3.01 cm

Answer 1

q1 = 4.98*10^-9 C

x1 , y1 = ( 0 , 0 )

q2 = -2.05*10^-9 C

(x2 , y2 ) = 3.97 cm , 0 = (0.0397 m , 0)

q3 = 6.04*10^-9 C

(x3 , y3) = 3.97 , 3.01 = ( 0.0397 m , 0.0301 m)

r13 = sqrt(x3^2+y3^2) = sqrt(0.0397^2 + 0.0301^2) = 0.0498 m

x component of force exerted by q1 on q3

F13x = k*q1*q3*x3/r13^3

F13x = 9*10^9*4.98*10^-9*6.04*10^-9*0.0397/0.0498^3

F13x = 8.7*10^-5 N

y component of force exerted by q1 on q3

F13y = k*q1*q3*y3/r13^3

F13y = 9*10^9*4.98*10^-9*6.04*10^-9*0.0301/0.0498^3

F13y = 6.6*10^-5 N

x component of force exerted by q2 on q3

F23x = 0

y component of force exerted by q2 on q3

F23y = -k*q2*q3/r23^2

F23y = -9*10^9*2.05*10^-9*6.04*10^-9/0.0301^2

F23y = -12.3*10^-5 N

x component of total force

Fx = F13x + F23x = 8.7*10^-5 N

y component of total force

Fy = F13y + F23y = -5.7*10^-5 N

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part C

magnitude = sqrt(Fx^2+Fy^2) = 10.4*10^-5 N

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part D

direction = tan^-1(Fy/Fx)

direction = 0.58 radians between F and +x axis