A thin fixed ring of radius $1 m$ has a positive charge $1 \times 1 0^{- 5} C$ uniformly distributed over it. A particle of mass $0.9 g m$ having a negative charge of $1 \times 1 0^{- 6} C$ is placed on the axis at a distance of $1 c m$ from the centre of the ring. Assuming that the oscillations has small amplitude, the time period of oscillations is:

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Answer 1

We know $E$ on the axis of the ring is:

$E = \frac{k q x}{( r ^{2} + x ^{2} ) ^{3 / 2}}$

For small amplitudes $r > > x$ :

$\Rightarrow E = \frac{k q x}{r ^{3}}$

We can see that force is always applied in the opposite direction of displacement i.e:

$F = \frac{- k q x}{r ^{3}} (1 \times 1 0^{- 6}) C$

$\Rightarrow a = \frac{- k q x ( 1 \times 1 0 ^{- 6} )}{m r ^{3}}$

We know for a particle in S. H. M: $a = - ω^{2} x$

$∴ ω^{2} = \frac{k q ( 1 \times 1 0 ^{- 6} )}{m r ^{3}} = \frac{9 \times 1 0 ^{9} \times 1 \times 1 0 ^{- 5} \times 1 \times 1 0 ^{- 6}}{0 \cdot 9 \times 1 0 ^{- 3} \times 1 ^{3}} = 100$

$\Rightarrow ω = 10$

Now, $T = \frac{2 π}{w} = \frac{2 \times 3 \cdot 1 4}{1 0} = 0.63 s$

Answer 2

We know

E

on the axis of the ring is:

E = \frac{k q x}{( r ^{2} + x ^{2} ) ^{3 / 2}}

For small amplitudes

r > > x

:

\Rightarrow E = \frac{k q x}{r ^{3}}

We can see that force is always applied in the opposite direction of displacement i.e:

F = \frac{- k q x}{r ^{3}} (1 \times 1 0^{- 6}) C

\Rightarrow a = \frac{- k q x ( 1 \times 1 0 ^{- 6} )}{m r ^{3}}

We know for a particle in S. H. M:

a = - ω^{2} x

∴ ω^{2} = \frac{k q ( 1 \times 1 0 ^{- 6} )}{m r ^{3}} = \frac{9 \times 1 0 ^{9} \times 1 \times 1 0 ^{- 5} \times 1 \times 1 0 ^{- 6}}{0 \cdot 9 \times 1 0 ^{- 3} \times 1 ^{3}} = 100

\Rightarrow ω = 10

Now,

T = \frac{2 π}{w} = \frac{2 \times 3 \cdot 1 4}{1 0} = 0.63 s