(25%) Problem 3: Four point charges ofequal magnitude Q-55 nC are placed on the

Question

(25%) Problem 3: Four point charges ofequal magnitude Q-55 nC are placed on the corners of a rectangle of sides DI-26 cm and D2 = // 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. -Q D2 Otheexpertta.com 20 % Part (a) Which of the following represents a free-body diagram for the charge on the lower left hand corner of the rectangle? Correet 20% Part (b) Calculate the horizontal component ofthe net force, in newtons on the charge which lies at the lower left corner of the triangle. 49.15 10 Grade Summary Deductions Potential 97% Submissions sin0 cotan) atan) cosh0 cos0 asin0 acotano sinh tanho cotanh0 ees Radians tano acoso Attempts remaining: J % per attempt) detailed view 0 END BACKSPACE CLEAR Submit Hint I give up! ints: .0% deduction per hint. Hints remaining- Feedback: deduction per fcedback 20% 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. -221.25 10- -2.213E-3 X Attempts Remain 20% Part (d) Calculate the magnitude of the net force, in newtons, on the charge located at the lower left corner of the rectangle. 20% Part e) Calculate the angle of the net force vector, measured counterclockwise rom the positive horizontal axis. Enter an angle betwe 80° and 180° 77.5 X Attempts Remain

Explanation / Answer

Electrostatic force is give by

F = k*q1*q2/r^2

force is attractive if both charge have same signs, and force is repulsive if both charge have same sign.

Now net force will be

Fnet = F1 + F2 + F3

F1 = force between charge at lower left and lower right charge (Force will be towards east)

F2 = force between charge at lower left and upper left charge (force will be towards south)

F3 = force between charge at lower left and upper right charge (Force will be towards north-east)

Now (all charges are same, So q1*q2 = q^2, where q = 55 nC)

F1 = F1x = k*q*q/r1^2

r1 = D1 = 26 cm = 0.26 m

F1x = 9*10^9*(55*10^-9)^2/0.26^2

F1x = (4.03*10^-4 N) i

F2 = F1y = k*q*q/r2^2

r2 = D2 = 11 cm = 0.11 m

F2y = 9*10^9*(55*10^-9)^2/0.11^2

F2y = 22.5*10^-4 N = -(22.5*10^-4 N) j

F3 = k*q*q/r^2

r = sqrt (D1^2 + D2^2) = sqrt (0.26^2 + 0.11^2)

r = 0.28 m

F3 = 9*10^9*(55*10^-9)^2/0.28^2

F3 = 3.47*10^-4 N

F3x = F3*cos A = 3.47*10^-4*(0.26/0.28)

F3x = (3.22*10^-4 N) i

F3y = F3*sin A = 3.47*10^-4*(0.11/0.28)

F3y = (1.36*10^-4 N) i

Now net force in x-direction will be

Fnet_x = F1x + F2x + F3x

Fnet_x = (4.03*10^-4 + 0 + 3.22*10^-4) i

Fnet_x = 7.25*10^-4 N

Now net force in y-direction will be

Fnet_y = F1y + F2y + F3y

Fnet_y = (0 - 22.5*10^-4 + 1.36*10^-4) i

Fnet_y = -21.14*10^-4 N

Now

Fnet = Fnet_x + Fnet_y

Fnet = 7.25*10^-4 i - 21.14*10^-4 j

|Fnet| = sqrt ((7.25*10^-4)^2 + (-21.14*10^-4)^2)

|Fnet| = 22.35*10^-4 N

Part D.

Direction = arctan (Fnet_y/Fnet_x)

= arctan (-21.14*10^-4/(7.25*10^-4))

= -71.07 deg

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