Question

Constants 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 Kand Eg-8.854x 10-12 C2/(N-m2) is the permittivity of free space Consider two point charges located on the x axis: one charge, q1-14.0 nC, is located at 11.715 m;the second charge, 92 32.0 nC, is at the origin (z = 0.0000)

Answer 1

Ans.

The charge q₃ is repelled by the charge q₂ at the origin whereas it is attracted to the charge q₁ at the location x₁. So the forces on q₃ due to q₁ and q₂ is in the negative x direction.

The magnitude of the net force on q₃ F = F₁ + F₂

F₁ = the magnitude of the force on q₃ due to q₁

F₂ = the magnitude of the force on q₃ due to q₂

_{The charge q_3 is repelled by the charge q_2 at the origin whereas it is attracted to the charge q_1 at the location x_1. So the forces on q_3due to q_1 and q_2 is in the negative x direction. The magnitude of the net force on q_3 F = F_1 + F_2 F_1 = the magnitude of the force on q_3 due to q_1 F_2 = the magnitude of the force on q_3 due to q_1 F_1 = k mod (q_1 q_3)/d^2_1 and F_2 = k mod (q_2 q_3)/s^2_2 q_1 = -14.0 * 10^-9 C, q_2 = 32.0 * 10^-9 C, q_3 = 45.0 * 10^-9 C The distance d_1 = mod (-1.715 - (-1.155)) = 0.56 m F = k mod (q_2 q_3)/d^2_1 + k mod (q_2 q_3)/d^2_2 Plugging the numbers to get the magnitude of the net force F F = 9 * 10^9 mod (-14.0 * 10^-9.middot 45.0 * 10^-9)/0.56^2 + 9 * 10^9 mod (32.0 * 10^-9 mid}

_and

Plugging the numbers to get the magnitude of the net force F

The net electric force on the charge q₃ = 2.780x10^-5 N in the negative x direction.