Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The total force on the particle with charge q is

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 =

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

= N cosine 60° + (-N cosine 60°) = 0

for the y-component

= -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 ( = 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.

dang that balls went far

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray