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The intensity of electric field required to balance a proton of mass $1.7 x 10^{- 27}$ kg and charge $1 . 6 x 10^{- 19}$ C is nearly

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$1 \times 10^{- 7} V / m$

$1 \times 10^{- 5} V / m$

$1 \times 10^{7} V / m$

$1 \times 10^{5} V / m$

answer is A.

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Detailed Solution

$Since qE = mg \Rightarrow E = \frac{mg}{q} = \frac{1.7 \times 10^{- 27} \times 9.8}{1.6 \times 10^{- 19}} = 10.0 \times 10^{- 8} = 1 \times 10^{- 7} V / m$

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