Question

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

• A. $2.304 \times 10^{- 10}N/C$
• B. $14.4V/m$
• C. $16V/m$
• D. $1.44 \times 10^{11}N/C$

Answer 1

Answer: (D)

$1.44 \times 10^{11}N/C$

Answer 2

$E = \frac{q}{4\pi\varepsilon_{0}r^{2}} = 9 \times 10^{9} \times \frac{1.6 \times 10^{- 19}}{(10^{- 10})^{2}} = 1.44 \times 10^{11}N/C$