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

\frac{40}{3} cm

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\frac{10}{3} cm

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20 cm

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10 cm

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Solution

The correct option is C $20 cm$
Distance from the smaller charge (magnitude)

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

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

 Hence, (c) is the correct answer.

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

The correct option is C 20 cmDistance from the smaller charge (magnitude) x=r√q2q1−1 =20√164−1=202−1=20 cm  Hence, (c) is the correct answer.

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

The correct option is C $20 cm$
Distance from the smaller charge (magnitude)

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

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

 Hence, (c) is the correct answer.