A) 7 m
B) 0.7 mm
C) 7cm
D) 0.7cm
Correct Answer: C
Solution :
Electric force, \[qE=ma\] \[\Rightarrow\] \[a=\frac{qE}{m}\] \[\therefore\] \[a=\frac{1.6\times {{10}^{-19}}\times 1\times {{10}^{3}}}{9\times {{10}^{-31}}}=\frac{1.6\times {{10}^{5}}}{9}\] \[u=5\times {{10}^{6}}\] and \[v=0\] \[\therefore\] From \[{{v}^{2}}={{u}^{2}}-2as\] \[\Rightarrow\] \[s=\frac{{{u}^{2}}}{2a}\] \[\therefore\] Distance, \[s=\frac{{{(5\times {{10}^{6}})}^{2}}\times 9}{2\times 1.6\times {{10}^{15}}}\simeq 7\,cm\]
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