Question

5)

a. A particle of charge 3 μC is located at the origin. A second particle of charge 3 μC is located at the coordinates (3.89,3.15) in cm. What is the magnitude of the electric force between the particles?

b. What are the x and y components of the electric force acting on the particle at the point (3.89,3.15)? [Enter the x component in the first box and the y component in the second box.] Answer 1 of 2: Answer 2 of 2:

Answer 1

In the triangle ABC , AB = 3.89 cm , BC = 3.15 cm

Using pythagorean theorem

AC = sqrt(AB² + BC²) = sqrt((3.89)² + (3.15)²) = 5.01 cm = 0.0501 m

so r = distance between the two charges = AC = 0.0501 m

magnitude of charge on each particle = Q₁ = Q₂ = 3 x 10^-6 C

Using coulomb's law , magnitude of force between the two particles is given as

F = k Q₁ Q₂/r²

F = (9 x 10⁹) (3 x 10^-6) (3 x 10^-6)/(0.0501)²

F = 32.3 N

b)

In triangle ABC

= tan^-1(BC/AB)

= tan^-1(3.15/3.89)

= 39 deg

X-component of the electric force between the two particles is given as

F_x = F Cos

F_x = (32.3) Cos39

F_x = 25.1 N

Y-component of the electric force between the two particles is given as

F_y = F Sin

F_y = (32.3) Sin39

F_y = 20.3 N