Question

Two fixed charges -2Q and Q are located at the points with coordinates (-3a, 0) and (+3a, 0),respectively, in the xy plane.

a. Show that all points in the xy plane where the electric potential due to the two charges is zero lie on a circle.Find its radius and the location of its center.

b. Give the expression V(x) at a general point on the x-axis and sketch the function V(x) on the whole x-axis.

c. If a particle of charge +q starts from rest at the center of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so.

Answer 1

Let P be a point in the xy plane with coordinates (x, y) at which the potential due to charges -2Q and +Q placed at A and B, respectively, be zero (Fig. S3.24).

$V_p=\frac{K(-2Q)}{\sqrt{(3a+x{)}^2+y^2}}+\frac{K(+Q)}{\sqrt{(3a-x{)}^2+y^2}}=0$

or $2\sqrt{(3a-x{)}^2+y^2}=\sqrt{(3a+x{)}^2+y^2}$

or $4\left[\right(3a-x{)}^2+y^2]=\left[\right(3a+x{)}^2+y^2]$

or $(x-5a{)}^2+(y-0{)}^2=(4a{)}^2$

This is the equation of a circle with center at (5a, 0) and radius 4a.Thus, c (5a, 0) is the center of the circle, where potential is zero.

b. Let us find potential at various points on the x-axis.For x > 3a,

$V=\frac{kQ}{x-3a}-\frac{k2Q}{x+3a}=\frac{kQ(9a-x)}{x^2-9a^2}$

We see that at x=9a, V=0.

For $-3a<x<3a,$

$V=\frac{kQ}{3a-x}-\frac{k2Q}{x+3a}=\frac{kQ3(x-a)}{9a^2-x^2}$

we see that at $x=a,V=0$

For x<-3a,

$V=\frac{kQ}{3a-x}+\frac{k(-2Q)}{-3a-x}=\frac{-kQ(9a-x)}{x^2-9a^2}$

We see that $V\ne 0$ for $x<-3a$ and $V$ is always negative for $x<-3a$.

c. Applying energy conservation, we get

$(KE+PE{)}_{center}=(KE+PE{)}_{circumference}$

or $0+K[\frac{Qq}{2a}-\frac{2Qq}{8a}]=\frac{1}{2}m{v}^2+K[\frac{Qq}{6a}-\frac{2Qq}{12a}]$

or $\frac{1}{2}m{v}^2=\frac{KQq}{4a}orv=\sqrt{\frac{KQq}{2ma}}=\sqrt{\frac{1}{4\pi {\epsilon }_0}\left(\frac{Qq}{2ma}\right)}$