Particle A of charge 3.00 x10 -4 C is at the origin, particle B of charge -6.00
ID: 1435816 • Letter: P
Question
Particle A of charge 3.00 x10-4 C is at the origin, particle B of charge -6.00 x 10-4 C is at (4.00 m, 0), and particle C of charge 1.00 x 10-4 C is at (0, 3.00 m). We wish to find the net electricl force on C. (a) What is the x component of the electricla force exerted by A on C? (b) What is the y component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B on C. (d) Calculate the x component of the force exerted by B on C. (e) Calculate the y component of the force exerted by B on C. (f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. (g) Similarly, find the y component of the resultant force vector acting on C. (h) Find the magnitude and direction of the resultant electric force acting on C.
Explanation / Answer
Distance between A and C is 3 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
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a)
Zero.
b)
y component of the electric force exerted by A on C
90*3*1 /3^2 = 30 N
c)
The magnitude of the force exerted by B on C.
90*6*1/5^2 = 21.6 N
d)
Angle of BC with the x axis 180 – and tan = ¾ and hence 180 – = 143.13°
the x component of the force exerted by B on C. is
21.6 cos 143.13 = -17.28 minus shows that it is toward left
e).
the y component of the force exerted by B on C.
21.6 sin 143.13 = 12.96 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 -17.28.
g)
Sum the two y components from part (a) and (d) = the resultant y component of the electric force acting along y -axis is 30 + 12.96 = 42.96 N
h)
Resultant of -17.28 and 42.96 is
[(-17.28)*2 + (42.96)^2] = 46.31 N
Direction is at angle (90 + ) from the x axis.
46.31 sin = 42.96 => = 68.07°
(90 + ) = 90 +68.07 = 158.07° from the x axis.
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