The figure above shows two point charges each of charge -Q.

a) Write an expression for the magnitude of the force on the charge placed on the x axis in terms of a, b, Q and fundamental constants.

b) Write an expression for the electrical potential in the origin due to both charges.

c) On the axis below sketch the horizontal component F

_{x}on the B charge as it is moving along the x axis.

d) The B charge is moved from the x axis to the y axis. Sketch the force F on the B charge as it is moved along the y axis.

Answer:

a) The distance between the particles is √(a

^{2}+ b

^{2}).

F = k[QQ/(a

^{2}+ b

^{2})]

F = kQ

^{2}/(a

^{2}+ b

^{2})

b) V = k(-Q)/a + k(-Q)/b = -kQ(1/a + 1/b)

c) F = k[QQ/(a

^{2}+ x

^{2})]

F

_{x}= k[Q

^{2}/(a

^{2}+ x

^{2})]cos(θ),

where θ is the angle betwen the force F and the x axis.

F

_{x}= k[Q

^{2}/(a

^{2}+ x

^{2})][x/√(a

^{2}+ x

^{2})] for positive values of x, and

F

_{x}= - k[Q

^{2}/(a

^{2}+ x

^{2})][x/√(a

^{2}+ x

^{2})] for negative values of x.

Using a graphic calculator we can sketch the shape of the force as a function of x.

d) F = kQ

^{2}/(y-a)

^{2}if y > a and,

F = - kQ

^{2}/(y-a)

^{2}if y < a and,

Barron's AP Physics B