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Question: How many electrons should be removed from a coin of mass 1.6 gm, so that it may float in electric fi...

How many electrons should be removed from a coin of mass 1.6 gm, so that it may float in electric field intensity 109 N/C directed upwards?

A

9.8 × 107

B

9.8 × 105

C

9.8 × 103

D

9.8 × 101

Answer

9.8 × 107

Explanation

Solution

qE = mg

neE = mg

n = mgeE\frac { \mathrm { mg } } { \mathrm { eE } } = 1.6×10−3×9.81.6×10−19×109\frac { 1.6 \times 10 ^ { - 3 } \times 9.8 } { 1.6 \times 10 ^ { - 19 } \times 10 ^ { 9 } }

n = 9.8 × 107