A) \[1.6\text{ }m/{{s}^{2}}\]
B) \[1.6\times {{10}^{18}}m/{{s}^{2}}\]
C) \[3.2\times {{10}^{18}}m/{{s}^{2}}\]
D) \[0.8\times {{10}^{18}}m/{{s}^{2}}\]
Correct Answer: B
Solution :
The force F on an electron (e) due to electric field (E) is given by \[F=eE\] ...(i) From Newtons law \[F=ma\] ...(ii) From Eqs. (i) and (ii), we get \[a=\frac{eE}{m}\] Given, \[E=9.1\times {{10}^{6}}N/C,\] \[e=1.6\times {{10}^{-19}}C,\] \[m=9.1\times {{10}^{-31}}kg\] \[a=\frac{1.6\times {{10}^{-19}}\times 9.1\times {{10}^{6}}}{9.1\times {{10}^{-31}}}\] \[\Rightarrow \] \[a=1.6\times {{10}^{18}}m/{{s}^{2}}\]You need to login to perform this action.
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