Question

Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal speeds of 4.0 m/s at an instant when the distance separating the two is equal to 25 cm. How far apart will they be when closest to one another?

Answer 1

Answer:

   r₁ = 20.5 cm

Explanation:

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

        Em₀ = U_g + U_e + K = $-G \frac{m_1m_2}{r} +k \frac{q_1q_2}{r} - 2 ( \frac{1}{2} m v^2)$

the two in the kinetic energy is because they are two particles

final point. When it is detained

        Em_f = U_g + U_e = $-G \frac{m_1m_2}{r_1} + k \frac{q_1q_2}{r_1}$

the energy is conserved

        Em₀ = em_f

the charges and masses of the two particles are equal

         $-G \frac{m^2}{r} + k \frac{q^2}{r} + m v^2 = - G \frac{m^2}{r_1} + k \frac{q^2}{r_1}$        

         

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

    -5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁  + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

      3.24 10⁻⁵ - 7.2 10⁻⁵ =  8.1 10⁻⁶ / r₁

      -3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

      r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

      r₁ = 2.045 10⁻¹ m

      r₁ = 20.5 cm