Question

Two concentric rings, one of radius '$a$' and other '$b$' have the charges $+q$ and $-{\left(2/5\right)}^{-3/2}q$ respectively, as shown in the figure. Find the ratio $b/a$, if a charge particle placed on the axis at $z=a$ is in equilibrium:

Answer 1

field due to a ring of charge $Q$ at a distance $x$ parallel to axis is $\frac{KQx}{(r^2+x^2{)}^{\frac{3}{2}}}$

field due to both rings is $zero$ as the particle is stationary

$\frac{Kqa}{(a^2+a^2{)}^{\frac{3}{2}}}+\frac{K(-q)(\frac{2}{5}{)}^{-\frac{3}{2}}a}{(a^2+b^2{)}^{\frac{3}{2}}}=0$

$\frac{Kqa}{(a^2+a^2{)}^{\frac{3}{2}}}=-\frac{K(-q)(\frac{2}{5}{)}^{-\frac{3}{2}}a}{(a^2+b^2{)}^{\frac{3}{2}}}$

$(a^2+a^2)=(a^2+b^2)\frac{2}{5}$

$4a^2=b^2$

$\frac{b}{a}=2$

answer is $2$