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Question

Physics Question on electrostatic potential and capacitance

An electron initially at rest, is accelerated through a potential difference of 200 volt, so that it acquires a velocity 8.4×106ms1{8.4 \times 10^6 m s^{-1}}, the value of e/me/m of electron will be

A

1.76×1011Ckg1{1.76 \times 10^{11} C kg^{-1}}

B

3.76×1012Ckg1{3.76 \times 10^{12} C kg^{-1}}

C

4×1011Ckg1{4 \times 10^{11} C kg^{-1}}

D

0.76×1011Ckg1{0.76 \times 10^{11} C kg^{-1}}

Answer

1.76×1011Ckg1{1.76 \times 10^{11} C kg^{-1}}

Explanation

Solution

Maximum kine KEmax=eVKE_{max} = eV
or 12mv2=eV\frac{1}{2} mv^2 = eV
=12(8.4×106)22×102=1.764×1011Ckg1= \frac{1}{2} \frac{(8.4 \times 10^6)^2}{2 \times 10^2} = 1.764 \times 10^{11} \,{ C \, kg^{-1}}