The electric force is exerted by a 40-nC charged particle located at the origin of a Cartesian coordinate system on a 20-nC charged particle located at (2.0 m, 2.0 m ).

Part A
Determine the direction of the force.
Determine the direction of the force.
Part B
Determine the magnitude of the force.
Express your answer with the appropriate units.

Question

30-01-2020
Physics

contestada

The electric force is exerted by a 40-nC charged particle located at the origin of a Cartesian coordinate system on a 20-nC charged particle located at (2.0 m, 2.0 m ).

Part A
Determine the direction of the force.
Determine the direction of the force.
Part B
Determine the magnitude of the force.
Express your answer with the appropriate units.

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cjmejiab cjmejiab · Answer 1 · 2020-01-31T05:04:15+01:00

To develop this problem we will make a diagram describing the problem. From there we will obtain the angle of the force, and quickly using Pythagoras the distance between the two points. Applying the electrostatic force formula we will find the magnitude of said force

PART A)

The direction of force is

[tex]tan \theta = \frac{2}{2}[/tex]

[tex]\theta = tan^{-1} \frac{2}{2}[/tex]

[tex]\theta = 45\°[/tex]

Therefore angle [tex]\theta[/tex] is 45° counter clockwise from +x axis

PART B)

Magnitude of force

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

Here

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Each charge

r = Distance

[tex]F = \frac{(9*10^{9})(40*10^{-9})(20*10^{-9})}{(\sqrt{8})^2}[/tex]

[tex]F = 900*10^{-9}N[/tex]