Question

A point charge of -0.5 µC is located at the origin. A second point charge of 12 µC is at x = 1 m, y = 0.5 m. Find the x and y coordinates of the position at which an electron would be in equilibrium.

Answer 1

Ok, so before I help you with the calculations, I want to tell something about my intuition when I see this problem.
1) The two point charges have different signs; one is positive, the other is negative.
2) A positive and a negative charge will be attracted to each other.
3) The forces acting on the two point charges (no electron yet!) will point along a line intersecting the positions of the two charges.
4) At some point (x,y) on this line we can place an electron which will give a resultant force of 0 N acting on the electron (equilibrium).

Answer 2

4) states that the vector force acting on the electron due to the positive charge, \[\vec{F_{+}}\], plus the vector force acting on the electron due to the negative charge, \[\vec{F_{-}}\] is equal to zero. Newton's first law, mathematically:
\[\Sigma\vec{F} = \vec{F_{+}} + \vec{F_{-}}\]

Answer 3

I meant \[\Sigma\vec{F} = \vec{F_{+}} + \vec{F_{-}} = \vec{0}\]