Respuesta :
The given question is incomplete. The complete question is as follows.
Two identical balls each have a mass of 35.0 grams and a charge of q = 3.50 \times 10^{-6}C[/tex]. The balls are released from rest when they are separated by a distance of 6.00 cm. What is the speed of each ball when the distance between them has tripled? Use k = [tex]9.00 \times 10^{9} Nm^{2}/C^{2}[/tex].
Explanation:
According to the conservation of energy, the formula will be as follows.
[tex]\frac{kq_{1}q_{2}}{r_{1}} = \frac{kq_{1}q_{2}}{(3r_{1})} + \frac{1}{2}mv^{2} + \frac{1}{2}mv^{2}[/tex]
or, [tex]\frac{kq_{1}q_{2}}{r_{1}}[1 - \frac{1}{3}] = mv^{2}[/tex]
Putting the given values into the above formula as follows.
[tex]\frac{kq_{1}q_{2}}{r_{1}}[1 - \frac{1}{3}] = mv^{2}[/tex]
[tex]\frac{9 \times 10^{9} \times (3.5 \times 10^{-6})^{2}}{0.09} \times \frac{2}{3} = \frac{35}{1000}v^{2}[/tex]
[tex]v^{2}[/tex] = 23.333
v = 4.83 m/s
Thus, we can conclude that speed of each ball when the distance between them has tripled is 4.83 m/s.
