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Q.

Two charges 4 × 10^–9 C and –16 × 10^–9C are separated by a distance 20 cm in air. The position of the neutral point from the small charge is

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a

20 cm

b

40/3cm

c

20/3cm

d

10/3cm

answer is C.

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Detailed Solution

$x = \frac{r}{\sqrt{\frac{q_{2}}{q_{1}}} - 1} = \frac{20}{\sqrt{\frac{16}{4}} - 1} = 20 cm$

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