Answer:
v = 1.15*10^{7} m/s
Explanation:
given data:
charge/ unit area[tex] = \sigma = 1.99*10^{-7} C/m^2[/tex]
plate seperation = 1.69*10^{-2} m
we know that
electric field btwn the plates is[tex] E = \frac{\sigma}{\epsilon}[/tex]
force acting on charge is F = q E
Work done by charge q id[tex] \Delta X =\frac{ q\sigma \Delta x}{\epsilon}[/tex]
this work done is converted into kinectic enerrgy
[tex]\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}[/tex]
solving for v
[tex]v = \sqrt{\frac{2q\Delta x}{\epsilon m}[/tex]
[tex]\epsilon = 8.85*10^{-12} Nm2/C2[/tex]
[tex]v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}[/tex]
v = 1.15*10^{7} m/s
