Capacity of an isolated sphere is increased $n$ times when it is enclosed by an earthed concentric sphere. The ratio of their radii is

\frac{n}{n - 1}

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\frac{n^{2}}{n - 1}

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\frac{2 n}{n + 1}

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\frac{2 n + 1}{n + 1}

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Solution

The correct option is B $\frac{n}{n - 1}$
Let the radius of isolated sphere be $a$ and that of earthed sphare be $b$ .
Capacitance of isolated sphere $C = 4 π ϵ_{o} a$
Capacitance of isolated sphere enclosed in earthed sphere $n C = \frac{4 π ϵ_{o} a b}{b - a}$
$∴$ $n C = \frac{C b}{b - a}$
Or $n b - n a = b$
Or $(n - 1) b = n a$
$⟹ \frac{b}{a} = \frac{n}{n - 1}$

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Similar questions

Q. The capacitance (C) for an isolated conducting sphere of radius (a) is given by

4 π ϵ_{o} a

. If the sphere is enclosed with an earthed concentric sphere. The ratio of the radii of the spheres being

\frac{n}{n - 1}

then the capacitance of such a sphere will be increased by a factor.

Q. If

L = lim n \to \infty n^{- n^{2}} {[(n + 1) (n + \frac{1}{2}) (n + \frac{1}{2^{2}}) \dots (n + \frac{1}{2^{n - 1}})]}^{n}

, then the value of

ln L

{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

is equal to

n \sum r = 0 r^{2} . (^{n} C_{r})^{2}

is equal to

$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2 n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$ is equal to [RPET 2000]

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Capacity of an isolated sphere is increased n times when it is enclosed by an earthed concentric sphere. The ratio of their radii is

The correct option is B nn−1Let the radius of isolated sphere be a and that of earthed sphare be b.Capacitance of isolated sphere C=4πϵoaCapacitance of isolated sphere enclosed in earthed sphere nC=4πϵoabb−a∴ nC=C bb−aOr nb−na=bOr (n−1)b=na⟹ ba=nn−1

Capacity of an isolated sphere is increased $n$ times when it is enclosed by an earthed concentric sphere. The ratio of their radii is