x.i5ss=\"studentBoxBorder container\"> Response 1 Two point charges, Q_1 = -6.8

Question

x.i5ss="studentBoxBorder container">Response 1 Two point charges, Q_1 = -6.8 muC and Q_2 = 1.6muC, are located between two oppositely charged parallel plates, asshown in the figure. The two charges are separated by a distance of x = 0.31 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 71000 N/C. Calculate the net electrostatic force on Q_1 and give its direction. Magnitude 11.01 x N Direction to the left to the right Two point charges, Q_1 = -6.8 muC and Q_2 = 1.6muC, are located between two oppositely charged parallel plates, asshown in the figure. The two charges are separated by a distance of x = 0.31 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 71000 N/C. Calculate the net electrostatic force on Q_1 and give its direction. Magnitude 11.0189E24 x N Direction to the left to the right Two point charges, Q_1 = -6.8 muC and Q_2 = 1.6muC, are located between two oppositely charged parallel plates, asshown in the figure. The two charges are separated by a distance of x = 0.31 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 71000 N/C. Calculate the net electrostatic force on Q_1 and give its direction. Magnitude 11.0189 x N Direction to the left to the right

Explanation / Answer

Here two forces are acting on the charge Q1. First one is force acting on Q1 due to the electricfield between the plates, and this force is to the left (negative xaxis) And second one is force of attraction acting on Q1 dueto Q2, and this force is to the right (positive x axis). (1) The electric force acting on Q1 due to the electricfield between the plates of the capacitor is                   F1= -EQ1                       = -(71000N/C)(6.8*10-6C)                       = -0.4828N 2) The force of attraction acting on Q1 due toQ2 is F2 =(9.0*109N.m2/C2)(6.8*10-6C)(1.6*10-6C)/ (0.31m)2                                                                               = 1.019 N Now the net force acting on Q1 is F = F1 +F2                                                        = -0.4828N + 1.019N                                                        = 0.536, to the right

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