A.) Find the x -component of the total force exerted on the third charge by the
ID: 2130485 • Letter: A
Question
A.) Find the x-component of the total force exerted on the third charge by the other two.
B.) Find the y-component of the total force exerted on the third charge by the other two.
C.) Find the magnitude of the total force acting on the third charge.
D.) Find the direction of the total force acting on the third charge.
Please show ALL work with detailed steps. Thank you.
A charge 5.02nC is placed at the origin of an xy-coordinate system, and a charge -2.02nC is placed on the positive x-axis at x = 3.96cm A third particle, of charge 5.95nC is now placed at the point x = 3.96cin , y = 2.97cm.Explanation / Answer
Force due to first charge,
F1=kq1q3/d^2 where q1=5.02 nc, q3=5.95 nc and d=(0.0396^2+0.0297^2)^0.5 m=0.0495 m
So F1=0.109711 mN along the line joining the first charge and the third charge away from origin
So the x, y components of F1 are as follows
F1x=F1cos? where ? is the angle made by the line joining the first charge and the third charge with the x-axis
Now tan?=(y-coordinate of third charge)/(x-coordinate of third charge)=2.97/3.96=0.75
So ?=tan-10.75=36.87 degrees
Thus F1x=F1cos?=0.109711 cos36.87 =0.0877 mN
and F1y=F1sin?=0.1094sin36.87=0.0658 mN
Force due to second charge,
F2=kq2q3/a^2 where a=0.0297 m
So F2=-0.1226 mN along the line joining the second and third charge, which is parallel to the y-axis. So no component of this force exists along the x-axis. The negative sign indicates that the force is attractive and hence directed towards the second charge
Thus F2=F2y=-0.1226 mN
Now net x-componet of all the forces on third charge is Fx=F1x only, i.e.
Fx=0.0877 mN
net y-component of all the forces on third charge is Fy=F1y+F2y=0.0658-0.11226 = -0.0568 mN
where mN=milli Newton
direction tan theta = -0.0568/0.08777 = -32.92
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.