Two concentric rings, one of radius R and total charge +Q and the second of radius 2 R and total charge -√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 the rings. At what distance from origin is the net electric field on z -axis zero?

Question

Two concentric rings, one of radius R and total charge +Q and the second of radius 2 R and total charge -√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 the rings. At what distance from origin is the net electric field on z -axis zero? (A) (R/2) (B) (R/√2) (C) (R/2 √2) (D) √2 R

Tardigrade Admin · Accepted Answer

Electric field at a point on z -axis distant r from origin is E=(1/4 π ε0)((Q r/(r2+R2)3 / 2)-(√8 Q r/(r2+4 R2)3 / 2))=0 Solving, we get r=√2 R

$\frac{R}{2}$

$\frac{R}{2}$

$\frac{R}{2 2}$

$2 R$

Electric field at a point on $z$ -axis distant $r$ from origin is $E = \frac{1}{4 π ε _{0}} (\frac{Q r}{( r ^{2} + R ^{2} ) ^{3/2}} - \frac{8 Q r}{( r ^{2} + 4 R ^{2} ) ^{3/2}}) = 0$
Solving, we get $r = 2 R$

2R​

2​R​

22​R​

2​R

Electric field at a point on z -axis distant r from origin is E=4πε0​1​((r2+R2)3/2Qr​−(r2+4R2)3/28​Qr​)=0 Solving, we get r=2​R

$\frac{R}{2}$

$\frac{R}{2}$

$\frac{R}{2 2}$

$2 R$

Electric field at a point on $z$ -axis distant $r$ from origin is $E = \frac{1}{4 π ε _{0}} (\frac{Q r}{( r ^{2} + R ^{2} ) ^{3/2}} - \frac{8 Q r}{( r ^{2} + 4 R ^{2} ) ^{3/2}}) = 0$
Solving, we get $r = 2 R$