Two positive point charges q are placed on the yaxis at y=a and y=a. A negative point charge Q is located at some point on the +xaxis.
A) Find the xcomponent of the net force that the two positive charges exert on Q. (Your answer should only involve k, q,Q , a, and the coordinate x of the third charge.) Fx=
B) Find the ycomponent of the net force that the two positive charges exert on Q. (Your answer should only involve k, q, Q, a, and the coordinate x of the third charge.) Fy=
C) What is the net force on the charge Q when it is at the origin (x=0)? F=
2 Answers

_________________________________
Two positive point charges q are placed on the yaxis at y=a and y=a.
A negative point charge Q is located at some point ‘x’ on the +xaxis.
As the magnitudes of charges are equal and their distances are equal , the magnitude of forces are equal
The xcomponents of two equal forces in same direction are added up
Edited
A) the xcomponent of the net force that the two positive charges exert on Q is Fx= – [ 2kqQx /(a^2+x^2)^(3/2)],the minus sign indicates Fx is in negative x – axis direction.
______________________________________
The y components of two equal forces in opposite direction cancel
B) The ycomponent of the net force that the two positive charges exert on Q is Fy = Zero
______________________________________
When Q is at origin , it is equally attracted by q at y=a and q at y = a,hence the two equal forces in opposite direction cancel
C) The net force on the charge Q when it is at the origin (x=0) is F = Zero
_____________________________________________

a.)Using Coulombs law of point charges, each point in the circle would exert a field of (kQ/n)/d where k is coulomb’s constant, Q/n is the magnitude of charge, and d is the distance between the particle and the point of measurement where d=(a^2+x^2)^.5 Since the charges are in a circle, all horizontal force is canceled and the vertical force is the resultant vector described by the fraction of the x/d (opposite over hypotenuse). When adding the forces of all the charges the magnitude of the electric field is therefore described by the equation E=xkQ/(d^2). b.) The equation is the same. This is because a the horizontal component of a charged portion of a uniform ring is canceled by the opposite portion of the ring since they are an equal distance from the position in which the electric field is being measured, just as are the opposing point charges described in part a.