Answer:
This galaxy belongs to the elliptical galaxy category. This is because it does not have spiral arms.
To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as
![F = \frac{kq_1q_2}{d^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Bd%5E2%7D)
Here
k = Coulomb's Constant
Charge of each object
d = Distance
Our values are given as,
![q_1 = 1 \mu C](https://tex.z-dn.net/?f=q_1%20%3D%201%20%5Cmu%20C)
![q_2 = 6 \mu C](https://tex.z-dn.net/?f=q_2%20%3D%206%20%5Cmu%20C)
d = 1 m
a) The electric force on charge
is
![F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}](https://tex.z-dn.net/?f=F_%7B12%7D%20%3D%20%5Cfrac%7B%20%289%2A10%5E9%20Nm%5E2%2FC%5E2%29%281%2A10%5E%7B-6%7D%20C%29%286%2A10%5E%7B-6%7D%20C%29%7D%7B%281%20m%29%5E2%7D)
![F_{12} = 54 mN](https://tex.z-dn.net/?f=F_%7B12%7D%20%3D%2054%20mN)
Force is positive i.e. repulsive
b) As the force exerted on
will be equal to that act on
,
![F_{21} = F_{12}](https://tex.z-dn.net/?f=F_%7B21%7D%20%3D%20F_%7B12%7D)
![F_{21} = 54 mN](https://tex.z-dn.net/?f=F_%7B21%7D%20%3D%2054%20mN)
Force is positive i.e. repulsive
c) If
, a negative sign will be introduced into the expression above i.e.
![F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}](https://tex.z-dn.net/?f=F_%7B12%7D%20%3D%20%5Cfrac%7B%289%2A10%5E9%20Nm%5E2%2FC%5E2%29%281%2A10%5E%7B-6%7D%20C%29%28-6%2A10%5E%7B-6%7D%20C%29%7D%7B%281%20m%29%5E%7B2%7D%7D)
![F_{12} = F_{21} = -54 mN](https://tex.z-dn.net/?f=F_%7B12%7D%20%3D%20F_%7B21%7D%20%3D%20-54%20mN)
Force is negative i.e. attractive
Under the assumption that the three rocks are dropped from the same height, they will hit the ground at the same speed. The gravity of Earth is virtually the same for any object that is small compared to the size of the Earth. The acceleration will change with the distance from the Earth, but this change is so small for the range of heights we work with (consider the range of heights from sea level to the tip of Mount Everest) that we can take the average value and assume it to be constant. This constant value of acceleration due to Earth's gravity is 9.80665m/s²
Because the objects fall under the same constant acceleration, they will hit the ground at the same speed.