Question

A particle of mass m and charge −q is projected from the origin with a horizontal speed v into a space having electric field of magnitude E directed downwards as shown in the figure. If the effect due to gravity can be neglected, then choose the correct statement(s) among the following.

Open in App

Solution

The correct option is **D** The horizontal and vertical components of acceleration are ax=0, ay=qEm

Since, there is no force acting on the mass in the x direction, Fx=0

∴ ax=0

Now, since the electric field acts downwards, the force acting on the particle in the y direction, Fy=qE

∴ Using Fy=may

ay=qEm

So, option (b) is the correct answer.

Now considering the second equation of motion, x=uxt+12axt2

∵ ux=v and ax=0

⇒x=vt .....(1)

Similarily, for y=uyt+12ayt2

∵ uy=0 and ay=qEm

y=12ayt2=12(qEm)t2 .....(2)

So, option (d) is the correct answer.

We, have from (1), t=xv

Substituting t=xv in (2), we get

y=12(qEx2mv2)

So, option (c) is the correct answer.

Now, kinetic energy KE is,

KE=12m(v2x+v2y)

where, vx=v and vy=ayt=qEmt

KE=12m(v2+q2E2t2m2)

So, option (a) is the incorrect answer.

Hence, options (b) , (c) and (d) are the correct answers.

Since, there is no force acting on the mass in the x direction, Fx=0

∴ ax=0

Now, since the electric field acts downwards, the force acting on the particle in the y direction, Fy=qE

∴ Using Fy=may

ay=qEm

So, option (b) is the correct answer.

Now considering the second equation of motion, x=uxt+12axt2

∵ ux=v and ax=0

⇒x=vt .....(1)

Similarily, for y=uyt+12ayt2

∵ uy=0 and ay=qEm

y=12ayt2=12(qEm)t2 .....(2)

So, option (d) is the correct answer.

We, have from (1), t=xv

Substituting t=xv in (2), we get

y=12(qEx2mv2)

So, option (c) is the correct answer.

Now, kinetic energy KE is,

KE=12m(v2x+v2y)

where, vx=v and vy=ayt=qEmt

KE=12m(v2+q2E2t2m2)

So, option (a) is the incorrect answer.

Hence, options (b) , (c) and (d) are the correct answers.

0

View More