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1. Electric Charges and Fields
easy
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
$14.4\,V/m$
C
$16\,V/m$
D
$1.44 \times {10^{11}}\,N/C$
Solution
(d) $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$
Standard 12
Physics