Answer:
D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q = [tex]\frac{kQq}{r^{2} }[/tex]
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
[tex]N_{x}[/tex] = N cosine 60° + (-N cosine 60°) = 0
for the y-component
[tex]N_{y}[/tex] = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero ([tex]N_{x}[/tex] = 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.
