Electric field is maximum due to a charged ring of radius R at a distance from circumference of the ring :

Question

Electric field is maximum due to a charged ring of radius R at a distance from circumference of the ring : (A) ( R /√2) (B) √(3/2) R (C) (√3/2) R (D) √2 R

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fe22e3ca2aacf980e64475c6021d6b76-.png" /> x=√R2+(R2/2)=√(3/2) R

Q. Electric field is maximum due to a charged ring of radius $R$ at a distance from circumference of the ring :

$\frac{R}{2}$

$\frac{3}{2} R$

$\frac{3}{2} R$

$2 R$

$x = R^{2} + \frac{R ^{2}}{2} = \frac{3}{2} R$

Q. Electric field is maximum due to a charged ring of radius R at a distance from circumference of the ring :

2​R​

23​​R

23​​R

2​R

x=R2+2R2​​=23​​R

Q. Electric field is maximum due to a charged ring of radius $R$ at a distance from circumference of the ring :

$\frac{R}{2}$

$\frac{3}{2} R$

$\frac{3}{2} R$

$2 R$

$x = R^{2} + \frac{R ^{2}}{2} = \frac{3}{2} R$