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A particle of mass 2 $μ$ g and carrying a charge of $1 \times 10^{- 9}$ C is moving directly towards a fixed positive point charge of magnitude $10^{- 8} C$ . When it is at a distance of 10 cm from the fixed point charge it has a velocity of $1 m s^{- 1}$ . At what distance from the fixed point charge will the particle come momentarily to rest? (in mm)

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answer is 47.

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Detailed Solution

By Law of Conservation of Energy

${(U + K)}_{\tan} = {(U + K)}_{\tan 1}$

$\Rightarrow \frac{1}{2} m v^{2} + \frac{Q q}{4 π ε_{0} r_{1}} = \frac{1}{2} m {(0)}^{2} + \frac{Q q}{4 π ε_{0} r_{2}}$

$\Rightarrow \frac{1}{2} m v^{2} = \frac{Q q}{4 π ε_{0}} (\frac{1}{r_{2}} - \frac{1}{r_{1}})$

$\Rightarrow \frac{1}{2} (2 \times 10^{- 6}) {(1)}^{2} = (10^{- 9}) (10^{- 8}) (9 \times 10^{9}) (\frac{1}{r_{2}} - \frac{1}{(\frac{10}{100})})$

$\Rightarrow \frac{100}{9} = \frac{1}{r_{2}} - 10$

$\Rightarrow \frac{1}{r_{2}} = \frac{190}{9} m$

$\Rightarrow r_{2} = 4.7 \times 10^{- 2} m$

$\Rightarrow r_{2} = 47 m m$

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