Two point charges -Q and +Q/√3 are placed in the xy-plane at the origin (0, 0) and a point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of radius R and potential V = 0 in the xy-plane with its center at (b, 0). All lengths are measured in meters.

Q.1 The value of R is ___ meter.
Q.2 The value of b is ___ meter.

Solution

$V_A=\frac {-kQ}{b-R}+\frac {kQ/\sqrt 3}{2-(b-R)}=0$

$\Rightarrow \frac {\sqrt 3}{b-R}=\frac {1}{R-b+2}=\frac {\sqrt 3+1}{2}$

$\Rightarrow b-R=\frac {2\sqrt 3}{\sqrt 3+1}=3-\sqrt 3$ ….(*)

$V_B=\frac {-kQ}{b+R}+\frac {kQ/\sqrt 3}{b+R-2}=0$

$\Rightarrow \frac {\sqrt 3}{b+R}=\frac {1}{b+R-2}=\frac {\sqrt 3-1}{2}$

$\Rightarrow b+R=\frac {2\sqrt 3}{\sqrt 3-1}=3+\sqrt 3$ ….(#)

From (*) & (#), b=3.00 & R=1.73