Two charged particles of masses m and 2m have charges +2q and +q respectively. They are kept in a uniform electric field E far away from each other an

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Two charged particles of masses m and 2m have charges +2q and +q respectively. They are kept in a uniform electric field E far away from each other and then allowed to move for the same time. The ratio of their kinetic energies is -
1. 8 ∶ 1
2. 1 ∶ 4
3. 1 ∶ 1
4. 4 ∶ 1

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Correct Answer - Option 1 : 8 ∶ 1

The correct answer is option 1) i.e. 8 ∶ 1

CONCEPT:

• The electric field is defined as the electric force per unit charge
•  A stationary charge q placed in an electric field intensity of E will experience an electrostatic force (F) given by

F = qE

• If a particle of mass m accelerates, from Newton's second law we know that the force experienced by this particle, F = ma
• For a charged particle the force experienced, F = qE

Therefore, qE = ma

⇒ Acceleration, a = qE/m

• Kinetic energy is the energy possessed by a moving object. It is given by

KE = $\frac{1}{2}$mv2

EXPLANATION:

Let the mass of two particles have mass m1 and m2 and charge q1 and q2.

m1 = m ; q1 = 2q

m2 = 2m ; q2 = q

We know that acceleration = velocity ÷ time ⇒ v = at

v1 = a1t = q1Et/m1

v2 = a2t = q2Et/m2

The kinetic energy of the particles will be as follows

KE1 = $\frac{1}{2}m_1v_1^2 = \frac{1}{2}(m)(\frac{(2q)E}{m} t)$

KE2 = $\frac{1}{2}m_2v_2^2 = \frac{1}{2}(2m)(\frac{qE}{2m} t)$

Ratio = $\frac{KE_1}{KE_2} =$$\frac{\frac{1}{2}(m)(\frac{2qE}{m} t)^2}{\frac{1}{2}(2m)(\frac{qE}{2m} t)^2} =\frac{2^2 / 1}{2/2^2} = \frac{4/1}{2/4} = \frac{16}{2} = \frac{8}{1}$

Therefore, the ratio of the kinetic energies is 8 : 1.