For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is :

Question

For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is : (A) (R/√5) (B) R (C) (R/√2) (D) R √2

Tardigrade Admin · Accepted Answer

Electric field on axis of ring E = (kQh/(h2+R2)3/2) for maximum electric field (dE/dh) = 0 ⇒ h = (R/√2)

Q. For a uniformly charged ring of radius $R$ , the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is :

$\frac{R}{5}$

$R$

$\frac{R}{2}$

$R 2$

Electric field on axis of ring
$E = \frac{k Q h}{( h ^{2} + R ^{2} ) ^{3/2}}$
for maximum electric field
$\frac{d E}{d h} = 0$
$\Rightarrow h = \frac{R}{2}$

Q. For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is :

5​R​

R

2​R​

R2​

Electric field on axis of ring E=(h2+R2)3/2kQh​ for maximum electric field dhdE​=0 ⇒h=2​R​

Q. For a uniformly charged ring of radius $R$ , the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is :

$\frac{R}{5}$

$R$

$\frac{R}{2}$

$R 2$

Electric field on axis of ring
$E = \frac{k Q h}{( h ^{2} + R ^{2} ) ^{3/2}}$
for maximum electric field
$\frac{d E}{d h} = 0$
$\Rightarrow h = \frac{R}{2}$