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

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

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(C) The electric field $E$ at a point $P$ on the axis of a uniformly charged ring at a distance $x$ from the center is given by:$E = \frac{kQx}{(R^2 + x^2)^{3/2}}$From the geometry of the right-angled triangle formed by the radius $R$,the axial distance $x$,and the distance $r$ from any point on the ring to point $P$,we have:$r^2 = R^2 + x^2$This implies $x = (r^2 - R^2)^{1/2}$.Substituting these 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}$

$A$ ring of radius $R$ is charged uniformly with a charge $+Q$. The electric field at a point on its axis at a distance $r$ from any point on the ring will be

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