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Q.

The distance between a proton and electron both having a charge $1 .6 \times 10^{- 19} coulomb$ , of a hydrogen atom is $10^{- 10} metre$ . The value of intensity of electric field produced on electron due to proton will be

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a

$14 .4 V / m$

b

$2 .304 \times 10^{- 10} N / C$

c

$16 V / m$

d

$1 .44 \times 10^{11} N / C$

answer is D.

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Detailed Solution

$E = \frac{q}{4 {πε}_{0} r^{2}} = 9 \times 10^{9} \times \frac{1 .6 \times 10^{- 19}}{{(10^{- 10})}^{2}} = 1 .44 \times 10^{11} N / C$

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