Question

The distance between a proton and electron both having a charge $1.6 \times 10^{-19} C$ of a hydrogen atom is $10^{-12}$ metre. The value of intensity of electric field produced on electron due to proton will be

• A. $14.4 \times 10^{14} N / C$
• B. $12.2 \times 10^{14} N / C$
• C. $11.4 \times 10^{14} N / C$
• D. $13.2 \times 10^{14} N / C$

Answer 1

Answer: (A)

$E =\frac{ Kq }{ r ^{2}}=\frac{9 \times 10^{9} \times 1.6 \times 10^{-19}}{\left(10^{-12}\right)^{2}}$
$E =9 \times 10^{9} \times 1.6 \times 10^{-19} \times 10^{24}$
$=14.4 \times 10^{14} N / C$