Answer:
a) attractiva, b) dF = , c) F =
, d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF =
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF =
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F =
F = k Q1 λ ()
we evaluate the integral
F = k Q₁ λ
F = k Q₁ λ
we change the linear density by its value
λ = Q2 / L
F =
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶)
F = -1.09 N
the sign indicates that the force is attractive
