Question

Problem 9: Four point charges of equal magnitude Q=55 nC are placed on the corners of a rectangle of sides D1=28 cm and D2 = 7 cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. Use a coordinate system fixed to the bottom left hand charge, with positive directions as shown in the figure

Part (a) Which of the following represents a free-body diagram for the charge on the lower left hand corner of the rectangle?

Part (b) Calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the triangle. 

Part (c) Calculate the vertical component of the net force, in newtons, on the charge which lies at the lower left corner of the triangle

Part(d) Calculate the magnitude of the net orce, in newtons, on the charge located at the lower left corner of the rectangle

Part (e) Calculate the angle of the net force vector, measured counterclockwise from the positive horizontal axis. Enter an angle between -180° and 180°.

Answer 1

b)

Force on the charge due to charge on the upper left corner:

F1 = k*Q^2/D2^2 = 9*10^9*(55*10^-9)^2/0.07^2

= 5.56*10^-3 N directed towards south

Force on the charge due to charge on the upper right corner :

F2 = k*Q^2/(D1^2 + D2^2)

= 9*10^9*(55*10^-9)^2/(0.28^2 + 0.07^2)

= 3.27*10^-4 N directed towards North east

Force on the charge due to charge on the lower right corner :

F3 = k*Q^2/D1^2

= 9*10^9*(55*10^-9)^2/(0.28^2)

= 3.47*10^-4 N directed towards right.

So, horizontal component of the force, Fx = F3 +F2*cos(14.03 deg)

= 3.47*10^-4 + 3.27*10^-4*cos(14.03 deg)

= 6.64*10^-4 N towards right

c)

Vertical component of the force, Fy = F1 - F2*sin(14.03 deg)

= 5.56*10^-3 - 3.47*10^-4*sin(14.03 deg)

= 5.48*10^-3 N towards south

d)

magnitude of net force , Fnet = sqrt((6.64*10^-4)^2 + (5.48*10^-3)^2)

= 5.52*10^-3 N <-------- answer

e)

direction = -atan(54.8/6.64)

= -83.1 deg <------ answer