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 center 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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{R}{2 \sqrt{2}}$

$\sqrt{2} R$

$\frac{R}{2}$

$\frac{R}{\sqrt{2}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Electric field at a point on z-axis distant r from origin is

$E = \frac{1}{4 π ε_{o}} (\frac{Q r}{{(r^{2} + R^{2})}^{3 / 2}} - \frac{\sqrt{8} Q r}{{(r^{2} + 4 R^{2})}^{3 / 2}}) = 0 \Rightarrow \frac{1}{{(r^{2} + R^{2})}^{3}} = \frac{8}{{(r^{2} + 4 R^{2})}^{3}} \Rightarrow \frac{1}{(r^{2} + R^{2})} = \frac{2}{(r^{2} + 4 R^{2})} \Rightarrow r^{2} + 4 R^{2} = 2 r^{2} + 2 R^{2}$

$\Rightarrow r = \sqrt{2} R$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET