You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Acceleration in 2D

A particle of...

Question

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)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

Similar questions

A particle of mass m and charge q is released from rest in a uniform electric field E. The kinetic energy attained by the particle after moving a distance x along the electric field is

Q. A particle having mass

m

and charge

q

is at rest. On applying a uniform electric field

E

on it,it starts moving. What is the kinetic energy when it travels a distance

x

in the direction of force?

Q. A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass? (Curves are drawn schematically and are not to scale).

Q. Assertion :If a charged particle is released from rest in a region of uniform electric and magnetic fields parallel to each other, it will move in straight line. Reason: The electric field exerts no force on the particle but the magnetic field does.

Q. A particle of mass

m

and charge

q

at rest is released in a uniform electric field between parallel planes of charge

+ q

and

- q

respectively. The particle accelerates towards the other place a distance

^{'} d^{'}

away. The speed at which it strikes the opposite plane is:

Join BYJU'S Learning Program

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