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A particle is unchanged and is thrown vertically upward from ground level with a speed of `5sqrt(5)` m//s in a region of space having uniform electric field As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of `13 m//s` Finally the particle is given a negative charge -q Ignoring air resistance determine the speed (in m//s) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height h. |
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Answer» Correct Answer - `0009` In uniform electric in vertical direction if (+ve) charge feels extra acceleration in downward direction then (-ve) charge will feel acceleration in upward direction. `v_(uncharged)=5sqrt(5)m//sec` `v=0, h=height` `v^(2)-u^(2)=-2(g) h` `-(5sqrt(5))=-2gh` `u_(q+)=13m//sec` `v=0 h = h` `v^(2)-u^(2)=2(g+F_(E)/m)h` `0-(13)^(2)=-2(g+(F_(E))/m)h` Let `u_(q-)=u(say)` `v=0 h=ht` `v^(2)-u^(2)=-2(g-(F^(E))/m)h` `-u^(2)=-2(g-(F_(E))/m)h,u=9m//sec` |
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