Particle A of charge 3.15 times 10^-4 C is at the origin, particle B of charge -

Question

Particle A of charge 3.15 times 10^-4 C is at the origin, particle B of charge -5.76 times 10^-4 C is at (4.00 m, 0), and particle C of charge 1.07 times 10^-4 C is at (0, 3.00 m). We wish to find the net electric force on C. (a) What is the x component of the electric force exerted by A on C? N (b) What is the y component of the force exerted by A on C? N (c) Find the magnitude of the force exerted by B on C. N (d) Calculate the x component of the force exerted by B on C. N (e) Calculate the y component of the force exerted by B on C. N (f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. N (g) Similarly, find the y component of the resultant force vector acting on C. N (h) Find the magnitude and direction of the resultant electric force acting on C. magnitude N direction degree counterclockwise from the +x-axis

Explanation / Answer

qA = 3.15*10^-4 C

qB = -5.76*10^-4 C

(x2 , y2) = (4 , 0)

qC = 1.07*10^-4 C

(x3 , y3) = (0,3)

force on qc due to qA

(a)

FAcx = 0

(b)

FAcy = k*qA*qc/y3^2 = 9*10^9*3.15*10^-4*1.07*10^-4/3^2 = 33.05 N

force on qc due to qA

r23 = sqrt(4^2+3^2) = 5 m

tantheteta = 3/4

theta = 36.8

(c)

Fbc = k*qB*qc/r23^2 = 9*10^9*5.76*10^-4*1.07*10^-4/5^2 = 22.2 N

(d)

Fbcx = Fbc*costheta = 17.8 N

(e)

Fbcy = -Fbc*sintheta = -13.3 N

(f)

Fx = 17.8 N

(g)

Fy = 33.05-13.3 = 19.75 N

(h)

F = sqrt(Fx^2+Fy^2) = sqrt(17.8^2+19.75^2) = 26.6 N

direction = tan^-1(Fy/Fx) = 48

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