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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

Two identical particles, each with a mass of 10 g and a charge of...

Two identical particles, each with a mass of 10 g10 \text{ g} and a charge of 2.0×10−4 C2.0 \times 10^{-4} \text{ C}, are held at a separation of 10 cm10 \text{ cm} and subsequently released. Determine the speeds of the particles when the separation becomes large.

Options

  1. A1200 m/s1200 \text{ m/s}
  2. B60 m/s60 \text{ m/s}
  3. C600 m/s600 \text{ m/s}
  4. D300 m/s300 \text{ m/s}

Correct answer

C. 600 m/s600 \text{ m/s}

Step-by-step solution

Initial potential energy of the system is given by Ui=14πϵ0q2rU_i = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{r}

Substituting the given values:

Ui=9×109×(2.0×10−4)20.1U_i = \dfrac{9 \times 10^9 \times (2.0 \times 10^{-4})^2}{0.1}

Ui=9×109×4.0×10−80.1=3600 JU_i = \dfrac{9 \times 10^9 \times 4.0 \times 10^{-8}}{0.1} = 3600 \text{ J}

When the separation becomes very large, the potential energy becomes zero and is entirely converted into the kinetic energy of the two particles.

By conservation of mechanical energy:

Ui=KfU_i = K_f

3600=12mv2+12mv2=mv23600 = \dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2 = mv^2

3600=(10×10−3)v23600 = (10 \times 10^{-3}) v^2

v2=360010−2=360000v^2 = \dfrac{3600}{10^{-2}} = 360000

v=600 m/sv = 600 \text{ m/s}
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