(C) The electric field $E$ at a distance $x$ on the axis of a charged ring is given by the formula:$E = \frac{KQx}{(R^2 + x^2)^{3/2}}$Given $x = \sqrt

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(C) The electric field $E$ at a distance $x$ on the axis of a charged ring is given by the formula:$E = \frac{KQx}{(R^2 + x^2)^{3/2}}$Given $x = \sqrt{8} R$,we substitute this into the formula:$E = \frac{KQ(\sqrt{8} R)}{(R^2 + (\sqrt{8} R)^2)^{3/2}}$$E = \frac{\sqrt{8} KQR}{(R^2 + 8R^2)^{3/2}}$$E = \frac{\sqrt{8} KQR}{(9R^2)^{3/2}}$$E = \frac{\sqrt{8} KQR}{27 R^3}$$E = \frac{\sqrt{8} KQ}{27 R^2}$Since $\sqrt{8} = 2\sqrt{2}$,the expression can also be written as $\frac{2\sqrt{2} KQ}{27 R^2}$.

Class 12 Physics Electric Charges and Fields Electric Field

Medium

What is the electric field at a distance $x = \sqrt{8} R$ from the center on the axis of a charged ring with charge $Q$ and radius $R$?

A
$\frac{\sqrt{2}}{27} \frac{KQ}{R^2}$
B
$\frac{\sqrt{3}}{27} \frac{KQ}{R^2}$
C
$\frac{\sqrt{8} KQ}{27 R^2}$
D
$\frac{KQ}{R^2}$

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Electric Charges and Fields

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MediumAIPMT 1995

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$A$ thin conducting ring of radius $R$ is given a charge $+Q.$ The electric field at the centre $O$ of the ring due to the charge on the part $AKB$ of the ring is $E.$ The electric field at the centre due to the charge on the part $ACDB$ of the ring is

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Three equal charges $+q$ each are placed at the three vertices of an equilateral triangle. The electric field at the centroid of the triangle is . . . . . . . ('$r$' is the length of the side of the triangle).

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What is the electric field at a distance $x = \sqrt{8} R$ from the center on the axis of a charged ring with charge $Q$ and radius $R$?

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The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\,N/C$ just above its surface is:

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