Question

A positively charged particle q is accelerated through a uniform electric field E starting from rest. If v0 is the velocity of the particle at the end of distance d, what is the velocity of the particle at the end of 2d?

Answer 1

Answer:

The work done by the potential difference gets stored as its kinetic energy.

∴

2

1

mv

2

=qV⇒v=(

m

2qV

)

1/2

(i)

When it enters the region xϵ[0,a), it experiences an electric field

E

=E

j

^

Time taken to cross the region: vt=a

⇒t=

v

a

=a(

2qV

m

)

1/2

(ii)

The distance travelled in y-direction during this time is

y=

2

1

m

qE

t

2

=

2

1

×

m

qE

×a

2

×

2qV

m

⇒y=

4

1

V

Ea

2

Hence, the particle meets the line x=a at point

(x,y)=(a,

4

1

V

Ea

2

)