Answer:
262 kN/C
Explanation:
If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:
v² = v0² + 2aΔS
Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:
(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013
16x10¹⁴ = 4x10¹⁴ + 0.026a
0.026a = 12x10¹⁴
a = 4.61x10¹⁶ m/s²
The electric force due to the electric field (E) is:
F = Eq
Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:
F = m*a
Where m is the mass, so:
E*q = m*a
The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:
E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶
E = 261,866.42 N/C
E = 262 kN/C
Answer:
electric field E = (1 /3 e₀) ρ r
Explanation:
For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.
The charge within our surface is
ρ = Q / V
Q ’= ρ V
'
The volume of the sphere is V = 4/3 π r³
Q ’= ρ 4/3 π r³
The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area
I E da = Q ’/ ε₀
E A = E 4 πi r² = Q ’/ ε₀
E = (1/4 π ε₀) Q ’/ r²
Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant
R = Q ’/ V’ = Q / V
How you want the solution depending on the density (ρ) and the inner radius (r)
Q ’= R V’
Q ’= ρ 4/3 π r³
E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³
E = (1 /3 e₀) ρ r
Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
i'm stuck on that question also
An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height
