Answer:
6.05 cm
Explanation:
The given equation is
2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)
The initial head velocity V₀ₓ =11 m/s
The final head velocity Vₓ is 0
The accelerationis given by =1000 m/s²
the stopping distance = x-x₀=?
So we can wind the stopping distance by following formula
2 (-1000)(x-x₀)=[]
x-x₀=6.05* m
=6.05 cm
Answer:
F= σ² L² /2ε₀
F = (L² ε₀/4π) ΔV² / r⁴
Explanation:
a) For this exercise we can use Coulomb's law
F = - k Q² / r²
where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates
Capacitance is defined by
C = Q / ΔV
Q = C ΔV
also the capacitance for a parallel plate capacitor is related to its shape
C = ε₀ A / r
we substitute
Q = ε₀ A ΔV / r
we substitute in the force equation
F = k (ε₀ A ΔV / r)² / r²
k = 1 / 4πε₀
F = ε₀ /4π L² ΔV² / r⁴4
F = L² ΔV² ε₀/ (4π r⁴)
F = (L² ε₀/4π) ΔV² / r⁴
b) Another way to solve the exercise is to use the relationship between the force and the electric field
F = q E
where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface
Ф = ∫E .dA = / ε₀
the plate have two side
2E A = q_{int} / ε₀
E = σ / 2ε₀
σ = q_{int} / A
substituting in force
F = q σ / 2ε₀
the charge total on the other plate is
q = σ A
q = σ L²
F= σ² L² /2ε₀
Answer:
See attached pictures.
Explanation:
See attachments for explanation.
Answer:
A). σ = 3.823 x
/N-
B). C/
C). J
Explanation:
A). We know magnitude of charge per unit area for a conducting plate is given by
where, E is resultant electric field = 1.2 x V/m
is permittivity of free space = 8.85 x
/N-
k is dielectric constant = 3.6
∴
= 3.6 x 8.85 x x 1.2 x
= 3.823 x
/N-
B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by
C/
C).
Area of the plate, A = 2.5
= 2.5 x
diameter of the plate, d = 1.8 mm
= 1800 m
∴ Total energy stored in the capacitor
J