A Positively charged oil droplet remains in the electric field between two horizontal plates separated by a distance 1 cm. If the charge on the drop is 3.2×10−19C and the mass of the drople is 1014 the potential difference the plates is
A
3125 V
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B
2532 V
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C
4978 V
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D
6162 V
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Solution
The correct option is A 3125 V FE=mg⇒q[V103]=10−14×10⇒3.2×10−9.(V10−2)=10−13V=10−133.2×10−17=100003.2=3125VHence,optionAiscorrectanswer.