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1. Electric Charges and Fields
easy
In a hydrogen atom, the distance between the electron and proton is $2.5 \times {10^{ - 11}}\,m$. The electrical force of attraction between them will be
A$2.8 \times {10^{ - 7}}\,N$
B$3.7 \times {10^{ - 7}}N$
C$6.2 \times {10^{ - 7}}N$
D$9.1 \times {10^{ - 7}}N$
Solution
(b)$F = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ – 19}} \times 1.6 \times {{10}^{ – 19}}}}{{{{(2.5 \times {{10}^{ – 11}})}^2}}} = 3.7 \times {10^{ – 7}}N$
Standard 12
Physics
