How many planets on the solar system?
2 answers:
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Answer:
The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.
Explanation:
Electron mobility, μ =
where
= Drift velocity
E = Electric field
Given that the electric field strength = 1.48 V/m,
Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is
4·μ =
Answer:
Explanation:
<em>The top 5 winners of NASA's design contest winner received amount of </em><em> </em>(Dollar One hundred thousand)<em>. </em>The respected amount was divided into 5 parts as there were 5 winners as well.
So, the amount was divided among the winners are as follows:
<em>Winner 1 ⇒ </em>
<em>Winner 2 ⇒ </em>
<em>Winner 3 ⇒ </em>
<em>Winner 4 ⇒ </em>
<em>Winner 5 ⇒</em>
Hence by adding all of the 5, we will get the answer, which is .
Answer:
Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C
Explanation:
a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,
The equation for the electric field produced by a point charge is
E = k q / r²
With r the distance between the point charge and the positive test charge
We look for each electric field
Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance
r² = x² + y²
E1 = k q / (x² + y²)
Particle 2. located at x = -24.9 m
r² = x² + y²
E2 = k q / (x² + y²)
We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.
In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find
Let's measure the angle from axis X
cos θ = CA / H = x / (x2 + y2) ½
E1x = E1 cos θ
E2x = = E1 cos θ
The resulting field
Ey = 0
Ex = E1x + E2x 2 E1x
Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)
Ex = kq 2x / ∛ (x² + y²)²
b) For this part we substitute the numerical values
Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂
Ex = 497.15 x / (x² + 0.062) ³/₂
Point where can the value of the electric field x = 38.1 cm = 0.381 m
Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂
Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2
Ex= 189.41 /0.0943
Ex = 2008 N / C
c) E = 1.00 kN / C = 1000 N / C
To solve this part we must find x in the equation
Ex = 497.15 x / (x² + 0.062) ³/₂
Let's use some arithmetic
Ex / 497.15 = x / (x² + 0.062) ³/₂
[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃
∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)
The roots of this equation are the solution to the problem,
For Ex = 1.00 kN / C = 1000 N / C
[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312
1.312 = (∛x² ) / (x² + 0.062)
1.312 (x² + 0.062) = ∛x²
1.312 X² - ∛x² + 1.312 0.062 = 0
1.312 X² - ∛x² + 0.0813 = 0
We need used computer
