Question

1. Three charges, +Q, +Q, and -Q are at the vertexes of an equilateral triangle a long on each side. Find the magnitude and direction of the force on one of the positive charges. Clearly set up and label. Show all steps leading to the final result.

2. An electron is projected from a distance of 0.500 m with a speed of 40.0 m/s toward another electron which is fixed in position. Determine the distance of closest approach where the moving electron stops and goes back. Clearly set up your equations and solve them.

charge on electron 1.602x10-19 C rest mass 9.109x10-31 kg Coulomb constant 8.988x109 N

Answer #1

**force on positive charge due to other
charges**

**due to positive charge : repulsive force**

**due to negative charge : attractive force**

**suppose on a triangle ABC the positive charge is at A
and other +q and -q charges are at B and C**

**so repulsive force is along BA and attractive force
along AC**

**both these force make and angle 120 degrees with each
other**

**Magnotude = kqq/a^2**

**net force in horizontal direction = 2kqqcos60/a^2 =
kqq/a^2 directed along right horizontal direction i.e parallel to
BC**

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Image:
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