Question

What is the net force exerted by these two charges on a third charge q3 = 45.0 nC placed between q1 and q2 at x3 = -1.105 m ?

Forces in a Three-Charge System Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is where K 1 and ATeo eo 8.854 x 10 12 C/(N ma) s the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 14.5 nC is located at z1 1.745 m the second charge, q2 33.5 nC is at the origin (z 0.0000). Part A What is the net force exerted by these two charges on a third charge ga 45.0nC placed between gi and g2 at 1.105 m Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures. Force on as 3.07.10

Answer 1

The diatance between chrge q₁=-14 nC and q₂=33.5 nC is 1.745 m .

another charge of 45 nC is placed between them at 1.105 m left from charge q₂ .

Distance between 45 nC charge and q_2, d=1.745 -1.105 =0.64 m

Force on 45 nC charge due to q1, F₁=K (14.5 $\times$ 10^-9 $\times$ 45 $\times$ 10^-9)/ (0.64)²   (attractive force , direction of force is towards left)

F₁=(9 $\times$ 10⁹)(14.5 $\times$ 10^-9 $\times$ 45 $\times$ 10^-9)/ (0.64)²

F₁=1.43 $\times$ 10^-5 N towards left.

Force on 45 nC charge due to q₂,  F₂=K (33.5 $\times$ 10^-9 $\times$ 45 $\times$ 10^-9)/ (1.105)²   (repulsive force ,direction of force is towards left)

F₂=(9 $\times$ 10⁹)(33.5 $\times$ 10^-9 $\times$ 45 $\times$ 10^-9)/ (1.105)²

F₂=1.11 $\times$ 10^-5 N towards left

Net force on 45 nC charge due to both of force is

F=F₁+F₂   (Both force in same direction)

F=1.43 $\times$ 10^-5 +1.11 $\times$ 10^-5 N

F=2.54 $\times$ 10^-5 N (towards left ) ANS