Question 457: Electrostats & Capacitance

Question

A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $$	imes$$ 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 :

Accepted Answer

$${v^2} - {u^2} = 2as$$
$$ \Rightarrow {0^2} - {200^2} = 2\left( {{{ - qE} \over m}} ight)(S)$$
$$ \Rightarrow  - {200^2} = 2\left[ {{{ - 40 	imes {{10}^{ - 6}} 	imes {{10}^5}} \over {100 	imes {{10}^{ - 6}}}}} ight][S]$$
$$ \Rightarrow S = {4 \over {2 	imes 4}}$$ m = 0.5 m

Answer

1 m

Answer

5 m

Answer

10 m

Answer

0.5 m

Electrostats & Capacitance

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 \muC and the initial velocity is 200 ms-$1, how much distance it will travel before coming to the rest momentarily :

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