A ring of radius R is uniformly charged. At what distance from the electric field will be maximum (on its axis).

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R/2

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\frac{R}{\sqrt{2}}

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\sqrt{2} R

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Solution

The correct option is C $\frac{R}{\sqrt{2}}$
Electric field due to ring on its axis is,
$ϵ = \frac{K Q x}{(x^{2} + R^{2})^{\frac{3}{2}}}$
We need to minimise the above expression, Therefore,
$\frac{d ϵ}{d x} = 0 \Rightarrow X = \frac{R}{\sqrt{2}}$ .

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A ring of radius R is uniformly charged. At what distance from the electric field will be maximum (on its axis).

The correct option is C R√2Electric field due to ring on its axis is,ϵ=KQx(x2+R2)32We need to minimise the above expression, Therefore,dϵdx=0⇒X=R√2.

The correct option is C $\frac{R}{\sqrt{2}}$
Electric field due to ring on its axis is,
$ϵ = \frac{K Q x}{(x^{2} + R^{2})^{\frac{3}{2}}}$
We need to minimise the above expression, Therefore,
$\frac{d ϵ}{d x} = 0 \Rightarrow X = \frac{R}{\sqrt{2}}$ .