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2. Electric Potential and Capacitance
easy
A proton has a mass $1.67 \times 10^{-27} \,kg$ and charge $+1.6 \times 10^{-19} \,C$. If the proton is accelerated through a potential difference of million volts, then the kinetic energy is ......... $J$
A
$1.6 \times 10^{-15}$
B
$1.6 \times 10^{-13}$
C
$1.6 \times 10^{-21}$
D
$3.2 \times 10^{-13}$
Solution
(b)
$\left(1.6 \times 10^{-19}\right)\left(10^6\right)=\frac{1}{2}\left(1.67 \times 10^{-27}\right) v^2$
$1.6 \times 10^{-13}=\frac{1.67}{2} \times 10^{-27} v^2= KE$
$KE =1.6 \times 10^{-13} \,J$
Standard 12
Physics
