Particle A of charge 2.97 Times 10^-4 C is at the origin, particle B of charge -
ID: 2001214 • Letter: P
Question
Particle A of charge 2.97 Times 10^-4 C is at the origin, particle B of charge -5.94 x 10^-4 C is at (4.00 m, 0), and particle C of charge 1.04 Times 10^-4 C is at (0, 3.00 m). We wish to find the net electric force on C. What is the x component of the electric force exerted by A on C? What is the y component of the force exerted by A on C? Find the magnitude of the force exerted by B on C. Calculate the x component of the force exerted by B on C. Calculate the y component of the force exerted by B on C. Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. Similarly, find the y component of the resultant force vector acting on Find the magnitude and direction of the resultant electric force acting on C.Explanation / Answer
a) Fx(A on C) = 0
b) Fy(A on C) = k*qA*qC/y^2
= 9*10^9*2.97*10^-4*1.04*10^-4/3^2
= 30.9 N
c) F(B on C) = k*qB*qC/y^2
= 9*10^9*5.94*10^-4*1.04*10^-4/(4^2 + 3^2)
= 22.24 N
d) Fx(B on C) = F(B on C)*cos(theta)
= 22.24*4/sqrt(3^2+4^2)
= 17.8 N
e) Fy(B on C) = F(B on C)*sin(theta)
= 22.24*3/sqrt(3^2+4^2)
= -13.34 N
f) Fnetx(on C) = 0 + 17.8
= 17.8
g) Fnety(on C) = 30.9 - 13.34
= 17.56 N
h) Fnet(on C) = sqrt(17.8^2 + 17.56^2)
= 25 N
direction : theta = tan^-1(Fy/Fx)
= tan^-1(17.56/17.8)
= 44.6 degrees counterclockwise from the +x axis.