A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int
1 answer:
Answer:
Explanation:
Let the charge on the large droplet be denoted as Q.
When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:
If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".
Applying volume conservation gives us:
Now, the potential for the smaller droplets is given as:
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Answer:
Baseball mass:
Explanation:
Circumference of a baseball is calculated using 2πr = 23 cm
Thus, the radius comes out to be 3.66 cm, which equals m
The mass density of the baseball matches that of a neutron or proton.
Proton mass = kg
Proton diameter = m
Proton radius = m
Volume of the baseball is
Now by substituting all values into the mass per unit volume equation for the baseball, we get:
Therefore, the baseball mass amounts to
- The greatest potential energy increase occurs when the charge travels north. This happens because the charge is negative, which means it gains potential energy when moving in the same direction as the field (in contrast, a positive charge moving along the field loses potential energy, converting it to kinetic energy). The potential energy gained is calculated as the charge multiplied by the distance moved:
- The next largest increase occurs as the charge moves east. Here, the change in potential energy is actually zero since the charge moves perpendicular to the field, traversing points with constant potential. Therefore, there is no variation in potential energy in this case:
- Finally, when the charge moves south, it experiences a reduction in potential energy. This is due to moving against the electric field, and since it is a negative charge, it loses potential energy in this direction, which transforms into kinetic energy. Thus, in this scenario:
- The next largest increase occurs as the charge moves east. Here, the change in potential energy is actually zero since the charge moves perpendicular to the field, traversing points with constant potential. Therefore, there is no variation in potential energy in this case:
- Finally, when the charge moves south, it experiences a reduction in potential energy. This is due to moving against the electric field, and since it is a negative charge, it loses potential energy in this direction, which transforms into kinetic energy. Thus, in this scenario: