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Two point charges 4µC and 9µC are separated by 50 cm. The potential at the point between them where the field has zero strength is

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9 × 10⁵V

4.5 × 10⁵V

9 × 10⁴V

Zero

answer is A.

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Detailed Solution

$x = \frac{50}{\sqrt{\frac{9}{4}} + 1} = 20 cm$ from 4µC At that point
$\begin{array}{l} V = 9 \times 10^{9} [\frac{4 \times 10^{- 6}}{20 \times 10^{- 2}} + \frac{9 \times 10^{- 6}}{30 \times 10^{- 2}}] \\ V = 4 .5 \times 10^{5} volt \end{array}$

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