A ring of radius R is charged with a charge Q | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The electric field $E$ at a distance $x$ from the center of a charged ring of radius $R$ on its axis is given by:$E = \frac{kQx}{(R^2 + x^2)^{3/2}

Vedclass · Accepted Answer

$\frac{KQ}{r^3} (r^2 - R^2)^{1/2}$

Vedclass · Answer

(C) The electric field $E$ at a distance $x$ from the center of a charged ring of radius $R$ on its axis is given by:$E = \frac{kQx}{(R^2 + x^2)^{3/2}}$From the geometry of the problem,the distance $r$ from the circumference to the point on the axis forms a right-angled triangle with the radius $R$ and the axial distance $x$. Thus,$r^2 = R^2 + x^2$,which implies $x = (r^2 - R^2)^{1/2}$.Substituting $x$ into the electric field formula:$E = \frac{kQ(r^2 - R^2)^{1/2}}{(r^2)^{3/2}}$$E = \frac{kQ(r^2 - R^2)^{1/2}}{r^3}$Therefore,the correct option is $C$.

$A$ ring of radius $R$ is charged with a charge $Q$. The electric field at a point on its axis at a distance $r$ from the circumference of the ring is:

Similar Questions

Two point charges $+8 q$ and $-2 q$ are located at $X=0$ (origin) and $X=L$ respectively. The net electric field due to these two charges is zero at point $P$ on the $X$-axis. The location of point $P$ from the origin is:

Two charges $+q$ and $-q$ are situated at a certain distance. At the point exactly midway between them:

The equation of an electric field line in the $xy$-plane is given by ${x^2} + {y^2} = 1$. What is true for a particle with unit positive charge initially at rest at the point $(1, 0)$ in the $xy$-plane?

Calculate the electric field at the origin due to an infinite number of charges as shown in the figure.

Two charges $+4e$ and $+e$ are at a distance $x$ apart. At what distance must a charge $q$ be placed from charge $+e$ so that it is in equilibrium?

Vedclass Test Series

Exam Paper Generator

Online Exam Module