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)

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Solution

The correct option is B

Using

$v^{2} - u^{2} = 2 a S$

$Here, u = 0 , s = x$

$A l s o, F_{e l e c t r i c} = m a$

$\Rightarrow q E = m a$

$\Rightarrow a = \frac{q E}{m}$

Substituting the values in $(i)$ we get

$v^{2} = \frac{2 q E}{m} . x$

Hence $v$ vs $x$ is parabolic in nature

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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)

The correct option is B Using v2−u2=2aS Here, u = 0 , s = x Also, Felectric=ma ⇒qE=ma ⇒a=qEm Substituting the values in (i) we get v2=2qEm.x Hence v vs x is parabolic in nature

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)

The correct option is B

Using

$v^{2} - u^{2} = 2 a S$

$Here, u = 0 , s = x$

$A l s o, F_{e l e c t r i c} = m a$

$\Rightarrow q E = m a$

$\Rightarrow a = \frac{q E}{m}$

Substituting the values in $(i)$ we get

$v^{2} = \frac{2 q E}{m} . x$

Hence $v$ vs $x$ is parabolic in nature