Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
![F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F_c%20-%20mg%5C%5C%5C%5CF_%7Bnet%7D%20%3D%20-4.548%20%2A10%5E%7B-3%7D%20-%20%281.1%2A10%5E%7B-3%7D%20%2A%209.8%29%5C%5C%5C%5CF_%7Bnet%7D%20%3D%20-15.328%2A10%5E%7B-3%7D%20%5C%20N)
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
![W = F_{net} *h\\\\W = 15.328 *10^{-3} * h](https://tex.z-dn.net/?f=W%20%3D%20F_%7Bnet%7D%20%2Ah%5C%5C%5C%5CW%20%3D%2015.328%20%2A10%5E%7B-3%7D%20%2A%20%20h)
W = K.E
![15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m](https://tex.z-dn.net/?f=15.328%2A10%5E%7B-3%7D%20%2Ah%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5C%5C%2015.328%2A10%5E%7B-3%7D%20%2Ah%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281.1%2A10%5E%7B-3%7D%29%284.8%29%5E2%5C%5C%5C%5C%2015.328%2A10%5E%7B-3%7D%20%2Ah%20%3D0.0127%5C%5C%5C%5Ch%20%3D%20%5Cfrac%7B0.0127%7D%7B15.328%2A10%5E%7B-3%7D%7D%5C%5C%5C%5C%20h%20%3D%200.827%20%5C%20m)
Therefore, the ball traveled 0.827 m