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Q.

Two charges $4 \times 10^{- 9} C$ and $16 \times 10^{- 9} C$ are separated by a distance 20cm in air. The position of the neutral point from the small charge is

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a

20 cm

b

$\frac{10}{3} cm$

c

$\frac{20}{3} cm$

d

$\frac{40}{3} cm$

answer is A.

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Detailed Solution

The distance of neutral point is given by

$x = \frac{r}{\sqrt{\frac{q_{2}}{q_{1}}} + 1}$ ,

$r = 20 cm$

$q_{1} = 4 \times 10^{- 9} C$

$q_{2} = 16 \times 10^{- 9} C$

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