Respuesta :
The given question is not complete. The complete question is as follows.
1. Calculate the potential energy of a system of two small spheres, one carrying a charge of 1.70 [tex]\mu C[/tex] and the other a charge of -4.40 [tex]\muC[/tex], with their centers separated by a distance of 0.240 m . Assume zero potential energy when the charges are infinitely separated.
2. Suppose that one of the spheres is held in place and the other sphere, which has a mass of 1.60 g , is shot away from it. What minimum initial speed would the moving sphere need in order to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it was infinitely distant from the fixed sphere).
Explanation:
1) . The given data is as follows.
k = [tex]9 \times 10^{9}[/tex]
d = 0.24 m
[tex]q_{1} = 1.70 \mu C = 1.70 \times 10^{-6} C[/tex]
[tex]q_{2} = 4.40 \mu C = 4.40 \times 10^{-6} C[/tex]
Hence, we will calculate the potential energy as follows.
Potential energy = [tex]\frac{kq_{1}q_{2}}{d}[/tex]
= [tex]\frac{9 \times 10^{9} \times 1.70 \times 10^{-6} C \times -4.40 \times 10^{-6} C}{0.24 m}[/tex]
= [tex]-280.5 \times 10^{-3}[/tex] J
Hence, value of the potential energy is [tex]-280.5 \times 10^{-3}[/tex] J.
2). According to the given situation, value of escaped kinetic energy will be equal to the potential energy as follows.
escaped K.E = -U
[tex]\frac{1}{2}mv^{2} = -(-280 \times 10^{-3}}[/tex]
[tex]\frac{1}{2}1.60 \times v^{2} = -(-280 \times 10^{-3}})[/tex]
[tex]0.8 \times v^{2} = 280 \times 10^{-3}[/tex]
[tex]v^{2} = 35 \times 10^{-2}[/tex]
v = 0.59 m/s
Therefore, minimum initial speed of the moving sphere is 0.59 m/s.
