A) 0.1 m
B) 0.2 m
C) 0.3 m
D) 0.4 m
Correct Answer: B
Solution :
(b) 0.2m \[F=qE=5\times {{10}^{-6}}\times 2\times {{10}^{5}}=1N\] Since, the particle is thrown against the field \[\therefore \,\,a=-F/m=-\frac{1}{{{10}^{-3}}}=-{{10}^{3}}\,m{{s}^{-2}}\] As \[{{v}^{2}}-{{u}^{2}}=2as\] \[\therefore \,\,{{0}^{2}}-{{\left( 20 \right)}^{2}}=2\times \left( -{{10}^{3}} \right)\times s\] Or \[s=0.2\,m\]You need to login to perform this action.
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