(a) What is the x component of the electric force exerted by A on C? N (b) What

Question

(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.

Explanation / Answer

given that

qA = 2.94*10^(-4) C

qB = -5.94*10^(-4) C

qC = 1.03*10^(-4) C

distance b/w A & C = 3 m

distance b/w A & B = 4 m

so distance b/w B & C = sqrt(3^2 + 4^2) = 5 m (using pythagorus theorum)

(a)

x component of the electric force exerted by A on C will be zero.

(b)

y component of the force exerted by A on C

(Fac)y = k*qA*qC / r^2

(Fac)y = 9*10^9*2.94*10^(-4)*1.03*10^(-4) / 3^2

(Fac)y = 30.28 N

(c)

magnitude of the force exerted by B on C

Fbc = 9*10^9*5.94*10^(-4)*1.03*10^(-4) / 5^2

Fbc = 22.02 N

(d)

internal angle b/w BC with the x axis = tan(teta) = 3/4

theta = 37 deg

x component of the force exerted by B on C

(Fbc)x = Fbc*cos(theta) = 22.02*cos37

(Fbc)x = 17.58 N

(e)

(Fbc)y = 22.02*sin37

(Fbc)y = 13.25 N

(f)

x component of the electric force acting on C

Fx = 17.58 N

(g)

Fy = 30.28 - 13.25 = 17.02 N

(h)

magnitude of the resultant electric force acting on C

F = sqrt(Fx^2 + Fy^2) = sqrt ((17.58)^2 + (17.02)^2 )

F = 24.47 N

direction of the resultant electric force acting on C

tan(theta) = Fy / Fx = 17.02 / 17.58

theta = 44.08 deg

answer

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