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A particle of mass `10^(-3)`kg and charge `5muC` is thrown at a speed of `20ms^(-1)` against a uniform electric field of strength `2xx10^(5)NC^(-1)`. The distance travelled by particle before coming to rest isA. 0.1 mB. 0.2 mC. 0.3 mD. 0.4 m |
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Answer» Correct Answer - B `F=qE=5xx10^(-6)xx2xx10^(5)=1N` Since, the particle is thrown against the field `therefore" "a=-F//m=-1/10^(-3)=-10^(3)ms^(-2)` `As" "v^(2)-u^(2)=2as` `therefore" "0^(2)-(20)^(2)=2xx(-10^(3))xxsors=0.2m` |
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