Question

Coulomb's law for the magnitude of the force FF between two particles with charges QQ and Q′Q′ separated by a distance dd is

|F|=K|QQ′|d2|F|=K|QQ′|d2,

where K=14πϵ0K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2)ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1q1 = -13.0 nCnC , is located at x1x1 = -1.670 mm ; the second charge, q2q2 = 39.0 nCnC , is at the origin (xx = 0).

What is (Fnet3)x(Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3q3 = 50.5 nCnC placed between q1q1 and q2q2 at x3x3 = -1.115 mm ?

Your answer may be positive or negative, depending on the direction of the force.

Express your answer numerically in newtons to three significant figures.

Answer 1

the electrostatic force constant is

k = 1/(4 Pi epsilon0) = 8.988 x 10^9

q3 is sitting between q1 and q2

since q3 is positive it will be attracted towards and q(which is negative)  and repelled by q2(which is positive)

distance between q3 and q1 is r1 = 1.670-1.115 = 0.555mm = 0.000555 m

distance between q3 and q2 is r2 = 1.115 mm = 0.00115 m

The net force will be in the negative x direction

F = k(q1 q3/r1^2+q2 q3/r2^2) =- 8.988 10^9(13 x 50.5/0.000555^2+39x 50.5/0.00115^2)x 10^-18

F = - 33.4 N

answered by: anonymous