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