It goes in the downward direction
The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
<h3>What is Columb's law?</h3>
The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Similar charges repel each other, whereas charges that are opposed attract each other.
Given data;
Electric force,F = 9 × 10 ⁻⁹ N
Distance between charges,d = 7 × 10⁻⁴ m
Chrge,q₁ = q₂ =q C
From Columb's law;
Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
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For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as
L=2.7*10^30w
<h3> What is its luminosity?</h3>
Generally, the equation for the luminosity is mathematically given as
L=4*\pi^2*b
Therefore
L=4*\pi^2*b
L=4* \pi *(2.83*10^{18})*2.7*10^{-8}
L=2.7*10^30
In conclusion, the luminosity
L=2.7*10^30w
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The most probable reason why the magnets won't stick on the refrigerator is that the body of the refrigerator and the magnets have like poles. If both have negative or both have positive poles facing each other, they will repel. In principle, magnets are attracted to opposite poles and like poles repel.
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.