Question

A charge of 3.20 μC is held fixed at the origin. A second charge of 3.20 μC is released from rest at the position (1.25 m, 0.570 m). a) If the mass of the second charge is 2.63 g , what is its speed when it moves infinitely far from the origin? (m/s)
b) At what distance from the origin does the second charge attain half the speed it will have at infinity? (m)

Answer 1

Answer:

Explanation:

Distance of second charge from origin d

d = $\sqrt{(1.25)^ 2+(0.57)^ 2}$

= 1.374 m

electric potential energy of system of charge

= 9 x 10⁹ x 3.2 x 3.2 x 10⁻¹² / 1.374

67.07 x 10⁻³ J

This energy will be converted into kinetic energy

1/2 x m v² = 67 x 10⁻³

v² = 2 x 67 x 10⁻³ / m

= 2 x 67 x 10⁻³/ 2.63 x 10⁻³

v = 7.13 m /s

b ) kinetic energy at this point = 67/4 x 10⁻³

= 16.75 x 10⁻³

Potential energy = 67x10⁻³ - 16.75 x 10⁻³ = 50.75  x 10⁻³

If r be the required distance

9 x 10⁹ x 3.2 x 3.2 x 10⁻¹² / r = 50.75 x 10⁻³

r = 1.87 m