Question

Three point charges are fixed on the x-axis: 6.0×10−6 C at x = −0.29 m, −4.0×10−6 C at the origin, and 4.9×10−6 C at x = 0.77 m. Find the electrostatic force acting on the charge at the origin due to the other two charges. Let a positive value indicate a force in the positive x direction, and a negative value indicate a force in the negative x direction. Answer in N.

Answer 1

let

Q1 = 6.0*10^-6 C at x = -0.29 m
Q2 = -4*10^-6 C at origin
Q3 = 4.9*10^-6 C at x = 0.77 m
d12 = 0.29 m
d23 = 0.77 m

magnitude of force exerted by Q1 on Q2, F1 = k*Q1*Q2/d12^2

= 9*10^9*6*10^-6*4*10^-6/0.29^2

= 2.57 N (towards -x axis)

magnitude of force exerted by Q3 on Q2, F3 = k*Q3*Q2/d23^2

= 9*10^9*4.9*10^-6*4*10^-6/0.77^2

= 0.2975 N (towards +x axis)

Net force on Q2, Fnet = -F1 + F2

= -2.57 + 0.2975

= -2.27 N <<<<<<<<-----------Answer

if answer is need in two significant figures, Fnet = -2.3 N