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Electrostatics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsElectrostatics3 Marks
Question
A particle carrying $5$ electrons starts from rest and is accelerated through a potential difference of $8900 V$. Calculate the $KE$ acquired by it in $MeV. [$Charge on electron $=1.6 \times 10^{-} \left.{ }^{19} C \right]$
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Answer

Data : $q =5 e , u =0, V =8900$
$V _{ l } e =1.6 \times 10^{-19} C$
$q =5\left(1.6 \times 10^{-19} C \right)=8 \times 10^{-19} C$
Initial $ KE = KE _{ i }=\frac{1}{2} mu ^2=0$
$\therefore \Delta KE = KE _{ f }- KE _{ i }= KE _{ f }$
$\Delta KE = qV$
$\therefore $ The final $ KE , KE _{ f }= qV$
$=\left(8 \times 10^{-19} C \right)(8900 V )$
$=7.12 \times 10^{-15} J$
$=\frac{7.12 \times 10^{-15}}{1.6 \times 10^{-19}} eV$
$=4.45 \times 10^4 eV$
$=\left(4.45 \times 10^{-2}\right) \times 10^6 eV$
$=4.45 \times 10^{-2} MeV $

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