JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 14)
A particle of mass m and charge q is released
from rest in a uniform electric field. If there is
no other force on the particle, the dependence
of its speed v on the distance x travelled by it
is correctly given by (graphs are schematic and
not drawn to scale)
_8th_January_Evening_Slot_en_14_1.png)
_8th_January_Evening_Slot_en_14_2.png)
_8th_January_Evening_Slot_en_14_3.png)
_8th_January_Evening_Slot_en_14_4.png)
Explanation
As F = qE
$$ \Rightarrow $$ ma = qE
$$ \Rightarrow $$ a = $${{qE} \over m}$$
Also we know, v2 = u2 + 2as
$$ \Rightarrow $$ v2 = 0 + 2$$\left( {{{qE} \over m}} \right)$$s
$$ \Rightarrow $$ v = $${\sqrt {{{2qE} \over m}} \sqrt x }$$
$$ \Rightarrow $$ ma = qE
$$ \Rightarrow $$ a = $${{qE} \over m}$$
Also we know, v2 = u2 + 2as
$$ \Rightarrow $$ v2 = 0 + 2$$\left( {{{qE} \over m}} \right)$$s
$$ \Rightarrow $$ v = $${\sqrt {{{2qE} \over m}} \sqrt x }$$
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