Question

two charged particles of masses 4m & m and having charges +q & +3q are placed in uniform electric field they are allowed to move for 2 sec find the ratio of their kinetic energy solution

Answer 1

The electrostatic force exerted on the charged particle can be given as:

F
=
q
E
.

Therefore, acceleration of the particle is given by:

a
=
F
m
=
q
E
m
.

The final velocity of the charged particle in time
t
is given by:

v
=
a
t
=
q
E
m
t
.

Then kinetic energy of the particle in time
t
is given by:

K
=
1
2
m
v
2
=
1
2
m
×
(
q
E
m
t
)
2
=
q
2
E
2
t
2
2
m
.

Therefore, if the electric field and time are kept constant, then:

K

∝

q
2
m
.

If we consider two charged particles of masses
m
1
and
m
2
and respective charges
q
1
and
q
2
, being accelerated in the same electric field and for the same interval of time, the ratio of their kinetic energies will be given by:

K
1
K
2
=
(
q
1
q
2
)
2
m
1
m
2
K
1
K
2
=
(
q
1
q
2
)
2
m
1
m
2
K
1
K
2
=
(
q
3
q
)
2
4
m
m
K
1
K
2
=
1
36