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A hollow spherical conductor of radius 4 cm carries a positive charge of $\frac{10}{9} \times 10^{- 10} C .$ At what distance (in cm) from centre of sphere should a positive point charge of $\frac{40}{9} \times 10^{- 10} C$ be placed so that the potential of the conductor becomes 75 V ? Assume potential to be zero at infinity.

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answer is 8.

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Detailed Solution

Net potential of the sphere is

$V = \frac{KQ}{R} + \frac{Kq}{x}$

$75 = \frac{(9 \times 10^{9}) (\frac{10}{9} \times 10^{- 10})}{4 \times 10^{- 2}} + \frac{(9 \times 10^{9}) (\frac{40}{9} \times 10^{- 10})}{x}$

On solving

$x = 8 cm$

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