Question

Particles of charge +65, +48, and −95 μC are placed in a line (Figure 1). The center one is L = 45 cm from each of the others.

A) Calculate the net force on the left charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left.

B) Calculate the net force on the center charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left.

C) Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right and a negative value if the force is directed to the left.

Answer 1

Force between the charges +65 AC and +48 that are at a distance of 45 cm from each other is replusive The magnitude of this force is given by (9x10° Nm2 /c) (65x10 C)(48x10 c) 138.6 N +65,+48 45x102 m Force between the charges +48 AC and-95 AC that are at a distance of 45 cm from each other is attractive The magnitude of this force is given by (9x10 Nm2/C) 48x10 c)(95x10 c) 45x 102 m -202.6 N +48,-95 Force between the charges +65 AC and-95 AC that are at a di stance of 90 cm from each other is attractive The magnitude of this force is given by (9 x10, Ntn18)(65 x 10-6 c)(95x10' С) 90x10-2 m 68.6 N +65.-95 Now the force on the charge at left (+65 μC) due to +48 is repulsive, acts toward left and due to-95 μc is attractive, acts toward right Therefore, the net force on + 65 μC charge is +65,+48 The force on the charge at center (+48 μ C) due to + 65 is repulsive, acts toward right and due to-95 μ0, is attractive, acts toward right. Therefore, the net force on +48 uC charge is 48F65e +FH5 138.6 N+202.6 N-341.2N +65,+43 +48,-95 The force on the charge at right -95 C) due to +48 uC is attractive, acts towar d left and due to 65 puC is attractive, acts towardleft Therefore, the net force on -95 AC charge is FsF6s 95- FHHR-95--68.6 N-202.6 N--271.2N