Particle A of charge 3.45 10^4 C is at the origin, particle B of charge 6.20 10^
ID: 1784468 • Letter: P
Question
Particle A of charge 3.45 10^4 C is at the origin, particle B of charge 6.20 10^4 C is at (4.14 m, 0) and particle C of charge 1.10 10^4 C is at (0, 3.56 m).
(a) What is the x-component of the electric 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 to obtain the resultant x-component of the electric force acting on C.
(g) Repeat part (f) for the y-component.
(h) Find the magnitude and direction of the resultant electric force acting on C.
Explanation / Answer
Distance between A and C is 3.56 m
Distance between B and C is sqrt(4.14^2 + 3.56^2) = 5.46 m
Formula for force = kq1q2/r^2
a) Zero.
A and C are both positive. So they repel each other
The force exerted by A on C is along the y-axis so has no x-component
so x-component of FAC = 0 N
b) y component of the electric force exerted by A on C
= 9*10^9*3.45*10^-4*1.1*10^-4/3.56^2
= 26.95 N
c) The magnitude of the force exerted by B on C.
FBC = 9*10^9*-6.2*10^-4*1.1*10^-4/5.46^2
= -20.6 N [ negative just means it's an attractive force]
d) Angle of BC with the x axis 180 – theta and
tan theta = 3.56/4.14
theta = arctan(3.56/4.14) = 40.69 degreee
so angle FBC makes with the positive x-axis is 270 + 40.69 = 310.69 degree
the x component of the force exerted by B on C. is
-20.6 sin 310.69 degree = -15.62 N minus shows that it is toward left
e). the y component of the force exerted by B on C.
= 20.6 cos 310.69 degree = 13.43 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 -15.62 N
g) Sum the two y components from part (a) and (d) = the resultant y component of the electric force acting along y -axis is 26.95 + 13.43 = 40.38 N
h) Resultant of -15.62 and 40.38 is
sqrt((-15.62)^2 + (40.38)^2) = 43.29 N
Direction of the resultant electric force on C = arctan (-15.62 / 40.38) = - 21.14 degree
right from the y-axis