Question

Two concentric rings, one of radius R and total charge +Q and the second of radius 2R and total charge $-\sqrt{8}Q$, lie in x - y plane (i.e. z = 0 plane). The common centre of rings lies at origin and the common axis coincides with the z-axis. The charge is uniformly distributed on both rings. At a distance $(x{)}^{y}R$ from origin the net electric field on z-axis zero. Find xy

Answer 1

Let the electric field be zero at a distance r on the z-axis
Electric field at a point on z-axis at distance r from origin,
$E=\frac{1}{4\pi {ϵ}_0}(\frac{Qr}{(r^2+R^2{)}^{\frac{3}{2}}}-\frac{\sqrt{8}Qr}{(r^2+4R^2{)}^{\frac{3}{2}}})$
At distance r from origin,
E = 0
$\Rightarrow E=\frac{1}{4\pi {ϵ}_0}(\frac{Qr}{(r^2+R^2{)}^{\frac{3}{2}}}-\frac{\sqrt{8}Qr}{(r^2+4R^2{)}^{\frac{3}{2}}})$
$\Rightarrow 0=\frac{1}{4\pi {ϵ}_0}(\frac{Qr}{(r^2+R^2{)}^{\frac{3}{2}}}-\frac{\sqrt{8}Qr}{(r^2+4R^2{)}^{\frac{3}{2}}})$
$\Rightarrow \frac{1}{(r^2+R^2{)}^{\frac{3}{2}}}=\frac{\sqrt{8}}{(r^2+2R^2{)}^{\frac{3}{2}}}$
$\Rightarrow r^2+4R^2=\left(8{)}^{1/3}\right(r^2+R^2)$
$\Rightarrow r^2+4R^2=2r^2+2R^2$
$\Rightarrow r=\sqrt{2}R$
$\Rightarrow x=2\text{\&}y=\frac{1}{2}$
$\Rightarrow xy=2\times \frac{1}{2}=1$
Hence, the value of xy is 1