Question

A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.

Answer 1

Answer:

a) attractiva, b) dF = $k \frac{Q_1 \ dQ_2}{dx}$, c)  F = $k Q_1 \frac{Q_2}{d \ (d+L)}$, 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 = $k \frac{Q_1 \ dQ_2}{dx}$

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 = $k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2$

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 = $k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}$

         F = k Q1 λ ($-\frac{1}{x}$)  

we evaluate the integral

        F = k Q₁ λ $(- \frac{1}{d+L} + \frac{1}{d} )$

        F = k Q₁ λ  $( \frac{L}{d \ (d+L)})$

we change the linear density by its value

      λ = Q2 / L

       F = $k Q_1 \frac{Q_2}{d \ (d+L)}$

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   $\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}$

       F = -1.09 N

the sign indicates that the force is attractive