A) 0.44mm
B) 0.44cm
C) 044m
D) None of these
Correct Answer: A
Solution :
Let the distance of electron from the negatively charged plate be 5, then Kinetic energy of electron =work done by electron \[KE=F\times s\] \[=Eq\times s\] \[K=\frac{\sigma }{{{\varepsilon }_{0}}}\times q\times s\] \[\therefore \] \[s=\frac{K{{\varepsilon }_{0}}}{\sigma q}\] \[=\frac{100\times 1.6\times {{10}^{-19}}\times 8.85\times {{10}^{-12}}}{2\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}}\] \[=50\times 8.85\times {{10}^{-6}}\] \[=442.50\times {{10}^{-6}}m\] \[=0.44\times {{10}^{-3}}m\] \[=0.44mm\]
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