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Case II. qa= qb= +6.5 C and qc = qd=-6.5 C. O qa Oqc (e) Due to symmetry the direction of the net force
Case II. qa= qb= +6.5 C and qc = qd=-6.5 C. O qa Oqc (e) Due to symmetry the direction of the net force is D. In the -y direction. N qb Oqd (d) In your notebook, draw the forces on q due to qar 9b, qc, and qd. Or use the result of of Homework: Charges on a Square Free Body Diagram. X Hint: For each force draw the x and y components. Some will add and some will cancel. (f) Calculate the magnitude of the force on the charge q, given that the square is 10.0 cm on a side and q = 1.7 C. Fnet
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