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Q.

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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a

9 × 105V

b

4.5 × 105V

c

9 × 104V

d

Zero

answer is A.

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Detailed Solution

x=5094+1=20cm from 4µC At that point
V=9×1094×10−620×10−2+9×10−630×10−2V=4.5×105 volt

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