Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ = ∫ E. dA = / ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B
Johann Strauss II
hope this helps
D. The atomic mass in amu is basically the number or nuclei since the mass of the electrons is negligible. For a given atom (element) the number of protons is fixed. Say the element has 10 protons. If the atomic weight is 14 atomic mass units (amu), you know that there are 4 neutrons, since both neutrons and protons are 1 amu each and there are 10 protons
I'm assuming the question is what is the robin's speed relative to to the ground...
Create an equation that describes its relative motion.
rVg = rVa + aVg
Substitute values.
rVg = 12 m/s [N] + 6.8 m/s [E]
Use vector addition.
| rVg | = √ | rVa |² + | aVg |²
| rVg | = √ 144 m²/s² + 46.24 m²/s²
| rVg | = √ 19<u>0</u>.24 m²/s²
| rVg | = 1<u>3</u>.78 m/s
Find direction.
tanФ = aVg / rVa
tanФ = 6.8 m/s / 12 m/s
Ф = 29°
Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]
The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
<h3>
Electric potential energy</h3>
When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.
The electric potential energy between the charges when the second charge is at point b is calculated as follows;
ΔU = -w
Ui - Uf = w
Uf = Ui - w
where;
Uf is the final potential energy
Ui is the initial potential energy
w is the work done by the force
Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)
Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J
Uf = 7.3 x 10⁻⁸ J
Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.
Learn more about electric potential energy here: brainly.com/question/14306881
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