Question

Two particles of masses m and 2m with charge 2q and 2q are placed in a uniform electric field E and allowed to move for the same time. The ratio of their kinetic energies will be

• A. 2 : 1
• B. 8 : 1
• C. 4 : 1
• D. 1 : 4

Answer 1

The correct option is A 2 : 1

Given,

The mass of the first particle is m

The mass of the second particle is 2 m

The charge on both the particles is 2 q

Electric field = E

We know,

When a charge is placed in an electric field then force acts on the charge

Force on both the charged particle $F=E\times 2q=2Eq$

We know,

$F=ma$

$Or,a=\frac{F}{m}$

So acceleration for the first and second particle is given as

$a_1=\frac{2qE}{m}$ and $a_2=\frac{2qE}{2m}$

Or,

$\therefore a_1=2a_2$

Now kinetic energy for the first particle
$K_1=\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}{m}_1(a_{1}t{)}^2$

Kinetic energy for the second particle
$K_2=\frac{1}{2}{m}_1{v}_2^2=\frac{1}{2}{m}_2(a_{2}t{)}^2$

Ratio of the kinetic energy
$\therefore \frac{K_1}{K_2}=\frac{m_1{a}_1^2}{m_2{a}_2^2}=\frac{m}{2m}\times \frac{4}{1}=\frac{2}{1}$

Hence, required ratio will be 2 : 1