Answer:
E = k Q₁ / r²
Explanation:
For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law
Ф = ∫ E .dA =
/ ε₀
where Ф the electric flow, qint is the charge inside the surface
To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone
R <r <R_a
for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.
E A = q_{int} /ε₀
The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field
q_{int} = Q₁
The surface area is
A = 4π r²
we substitute
E 4π r² = Q₁ /ε₀
E = 1 / 4πε₀ Q₁ / r²
k = 1/4πε₀
E = k Q₁ / r²
Answer:
e) upwrad z axis
Explanation:
To know the direction of the force we use the right hand rule.
The thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm of the hand in the direction of the force, this is for a positive charge if the charge is negative the force is in the opposite direction of The palm of the hand.
Let's apply to our case.
The thumb is in the x direction, the fingers in the vertical direction and since the electron has a negative charge, the force is on the z axis (perpendicular to the blade, coming out)
In general, in the nomenclature of the cardinal points the positive x-axis is the East, the positive y-axis is the North. Therefore the answer must be up on the z axis
The northern part of the united states does not have the same temperature climate the tropical plants need to survive but they could grow if the area they were growing in was a controlled climate. if I am right please mark brainlyest.
Frequency of any wave = (speed) / (wavelength)
Frequency = (3 x 10⁸ m/s) / (3.2 x 10⁻⁹ m) = <em>9.375 x 10¹⁶ Hz</em>
= 93,750,000 GHz
The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s
The distance between the two objects is
x = x₁ - x₂
= 4t + 161 - 48t + 4t²
x = 4t² - 44t + 161
Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
= 4[ (t - 5.5)² - 5.5² ] + 161
x = 4(t-5)² + 40
Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.
Answer: The distance of the closest approach is 40 m.