JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 16)
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $$\times$$ 105 NC$$-$$1. If the charge on the particle is 40 $$\mu$$C and the initial velocity is 200 ms$$-$$1, how much distance it will travel before coming to the rest momentarily :
1 m
5 m
10 m
0.5 m
Explanation
$${v^2} - {u^2} = 2as$$
$$ \Rightarrow {0^2} - {200^2} = 2\left( {{{ - qE} \over m}} \right)(S)$$
$$ \Rightarrow - {200^2} = 2\left[ {{{ - 40 \times {{10}^{ - 6}} \times {{10}^5}} \over {100 \times {{10}^{ - 6}}}}} \right][S]$$
$$ \Rightarrow S = {4 \over {2 \times 4}}$$ m = 0.5 m
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