A charge 5.03 nC is placed at the origin of an xy -coordinate system, and a char
ID: 1587053 • Letter: A
Question
A charge 5.03 nC is placed at the origin of anxy-coordinate system, and a charge -1.95 nC is placed on the positive x-axis at x = 3.95 cm . A third particle, of charge 5.99 nC is now placed at the point x = 3.95 cm , y = 3.03
a)Find the y-component of the total force exerted on the third charge by the other two.(in newtons)
b)Find the magnitude of the total force acting on the third charge.(in newtons)
c)Find the direction of the total force acting on the third charge.
cm
= radians between F and +x-axisExplanation / Answer
let
q1 = 5.03 nc
q2 = -1.95 nc
q3 = 5.99 nc
a) F13 = k*q1*q3/r13^2
= 9*10^9*5.03*10^-9*5.99*10^-9/(0.0395^2 + 0.0303^2)
= 1.1*10^-4 N
F23 = k*q2*q3/r23^2
= 9*10^9*1.95*10^-9*5.99*10^-9/(0.0303^2)
= 1.145*10^-4 N
Fnety = F13*sin(theta) - F23
= 1.1*10^-4*3.03/sqrt(3.03^2 + 3.95^2) - 1.145*10^-4
= -4.75*10^-5 N <<<<<------Answer
b) Fnetx = F13*cos(theta)
= 1.1*10^-4*3.95/sqrt(3.03^2 + 3.95^2)
= 8.73*10^-5 N
Fnet = sqrt(Fnetx^2 + Fnety^2)
= sqrt(4.75^2 + 8.73^2)*10^-5
= 9.94*10^-5 N <<<<<------Answer
c) theta = tan^-1(Fnety/Fnetx)
= tan^-1(-4.75/8.73)
= - 0.498 radians <<<<------Answer
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