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

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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:

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