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 z-axis. The charge is uniformly distributed on both rings. At what distance from origin is the net electric field on z-axis zero?

• A. $\frac{R}{2}$
• B. $\frac{R}{\sqrt{2}}$
• C. $\frac{R}{2\sqrt{2}}$
• D. $\sqrt{2}R$

Answer 1

Answer: (D)

$\sqrt{2}R$

Answer 2

The electric field on the axis of a uniformly charged ring of radius $r$ and charge $q$ at a distance $z$ from the center is given by $E_z = \frac{1}{4\pi\epsilon_0} \frac{qz}{(r^2 + z^2)^{3/2}}$. For the first ring with radius $R$ and charge $+Q$, the field is $E_1(z) = k \frac{Q z}{(R^2 + z^2)^{3/2}}$. For the second ring with radius $2R$ and charge $-\sqrt{8}Q$, the field is $E_2(z) = k \frac{-\sqrt{8}Q z}{((2R)^2 + z^2)^{3/2}} = k \frac{-\sqrt{8}Q z}{(4R^2 + z^2)^{3/2}}$. Setting the net electric field $E_{net}(z) = E_1(z) + E_2(z)$ to zero (for $z \neq 0$), we get $\frac{1}{(R^2 + z^2)^{3/2}} = \frac{\sqrt{8}}{(4R^2 + z^2)^{3/2}}$. Raising both sides to the power of $2/3$ gives $\frac{4R^2 + z^2}{R^2 + z^2} = (8^{1/2})^{2/3} = 8^{1/3} = 2$. Solving for $z^2$: $4R^2 + z^2 = 2(R^2 + z^2) \implies 4R^2 + z^2 = 2R^2 + 2z^2 \implies 2R^2 = z^2$. Thus, the distance from the origin is $z = \sqrt{2}R$.