Particle A of charge 3.45 times 10^-4 C is at the origin, particle B of charge 6

Question

Particle A of charge 3.45 times 10^-4 C is at the origin, particle B of charge 6.20 times 10^-4 C is at (3.60 m, 0) and particle C of charge 1.10 times 10^-4 C is at (0, 3.50 m). 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 to obtain the resultant x-component of the electric force acting on C. Repeat part (f) for the y-component. Find the magnitude and direction of the resultant electric force acting on C. magnitude direction counterclockwise from the +x-axis

Explanation / Answer

Distance between A and C is 3.5 m
Distance between B and C is 5 m
Formula for force = 9e9 q1q2/r^2

Use 9e9*ax1e-4*b*1e-4 = 90 ab
===========================
a)
Zero.

b)
y component of the electric force exerted by A on C
90*3.45*1.1 /3.5^2 = 27.88 N

c)
The magnitude of the force exerted by B on C.
90*6.2*1.1/5^2 = 24.55 N

d)
Angle of BC with the x axis 180 – and tan = 3.5/3.6 and hence 180 – = 135.81°
the x component of the force exerted by B on C. is
24.55 cos 135.81 = -17.60 minus shows that it is toward left

e).
the y component of the force exerted by B on C.
21.55 sin 135.81 = 17.11 N

f)
Sum the two x components from part (a) and (d) = the resultant x component of the electric force acting along x -axis is -17.60

g)
Sum the two y components from part (a) and (d) = the resultant y component of the electric force acting along y -axis is 27.88 + 17.11= 45.99 N

h)

Resultant of -17.60 and 45.99 is
[(-17.60)*2 + (45.99)^2] = 49.24 N
Direction is at angle (90 + ) from the x axis.

49.24 sin = 45.99 => = 69.06°

(90 + ) = 90 +68.07 = 159.07° from the x axis.

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