Answer:
at y=6.29 cm the charge of the two distribution will be equal.
Explanation:
Given:
linear charge density on the x-axis, 
linear charge density of the other charge distribution, 
Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.
Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.
<u>we know, the electric field due to linear charge is given as:</u>
where:
linear charge density
r = radial distance from the center of wire
permittivity of free space
Therefore,
∴at y=6.29 cm the charge of the two distribution will be equal.
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
From Alyssa's point of view, the water balloon is at first at rest and then gets thrown with a velocity of 23m/s. Therefore the balloon will have a speed of 23m/s for Alyssa.
At the same time, Naya is watching, and she sees the balloon at the beginning moving at a speed of 14m/s along with Alyssa, and then pushed forward of other 23m/s. Therefore, from her point of view, the balloon will have a speed of 14+23 = 37m/s.
Hence, the correct answer is <span>D) The speed of the balloon is 23 m/s for Alyssa and 37 m/s for Naya.
</span>
Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
