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ush19ky.theexperta.com The Expert TA | Human-like Grading, Automated! Chegg Study | Guided Solutions and Study Help | Chegg.com w T Begin Date ue ate: 00 PM End Date: 914120I IT:59:00 P (7%) Problem 7: Four point charges of equal magnitude Q = 35 nC are placed on the corners of a rectangle of sides D1 = 17 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 Assignment Status Click here for detailed view ©theexpertta.com Problem Status 20% Part (a) Which of the following represents a free-body diagram for the charge on the lower left hand corner of the rectangle? > 20% 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 Completed Completed Completed Completed 6 Completed Grade Summary 0% 100% Potential Submissions Attempts remaining: S (0% per attempt) detailed view sinO tan() | cos) cotanasin acos) atan) acotan sinh) cosh)tanh0cotanh0 Degrees Radians Partial Completed Completed Completed Completed 12 Completed 3 Completed 0% 0% Submit Hint I give up! 4 Partial 15 Partial Hints: 1 % deduction per hint. Hints remaining: 3. Feedback:-deduction per feedback. 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 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 from the positive horizontal axis. Enter an angle between -180° and 180°Explanation / Answer
partB] Fx = kq^2/D1^2 + kq^2/[D1^+D2^2] * D1/sqrt(D1^+D2^2)
= 9e9*35e-9^2/0.17^2 + 9e9*35e-9^2/(0.17^2+0.07^2)* 0.17/sqrt(0.17^2+0.07^2)
= 0.000683 N
partC] Fy= -kq^2/D2^2 + kq^2/[D1^+D2^2] * D1/sqrt(D1^+D2^2)
= -9e9*35e-9^2/0.07^2 + 9e9*35e-9^2/(0.17^2+0.07^2)* 0.07/sqrt(0.17^2+0.07^2)
= -0.002126 N
partD] net force = sqrt(0.002126^2+0.000683^2)
= 0.002233 N
partE] angle = 360 - arctan(0.002126/0.000683)
= -72.2 degree
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