Solveeit Logo
Login

Question

Physics Question on Electric charges and fields

Under the action of a given coulombic force the acceleration of an electron is 2.5×1022 m s−22.5 \times 10^{22}\, m \,s^{-2}. Then the magnitude of the acceleration of a proton under the action of same force is nearly

A

1.6×1019 m s−21.6 \times 10^{19}\, m \,s^{-2}

B

9.1×1031 m s−29.1 \times 10^{31}\, m \,s^{-2}

C

1.5×1019 m s−21.5 \times 10^{19}\, m \,s^{-2}

D

1.6×1027 m s−21.6 \times 10^{27}\, m \,s^{-2}

Answer

1.5×1019 m s−21.5 \times 10^{19}\, m \,s^{-2}

Explanation

Solution

The acceleration due to given coulombic force FF is
a=Fma=\frac{F}{m} or a∝1m ...(i)a\propto \frac{1}{m}\,...\left(i\right)
∴apae=memp\therefore \frac{a_{p}}{a_{e}}=\frac{m_{e}}{m_{p}} , where mem_{e} and mpm_{p} are masses of electron and proton respectively
ap=aememp=(2.5×1022 m s−2)(9.1×10−31 kg)(1.67×10−27 kg)a_{p}=\frac{a_{e}m_{e}}{m_{p}}=\frac{\left(2.5\times10^{22}\,m\,s^{-2}\right)\left(9.1\times10^{-31}\,kg\right)}{\left(1.67\times10^{-27}\,kg\right)}
=13.6×1018 m s−2≈1.5×1019 m s−2=13.6\times10^{18}\,m\,s^{-2}\approx1.5\times10^{19}\,m\,s^{-2}