Answer:
Light provides brightness to see and also light carries energy
Answer:
exponential
Explanation:
type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is
y = A ×
× cos(ωt + ∅ ) ..................................... ( 1 )
here function A ×
is amplitude
as per equation ( 1 )it is exponential
so that we can say that amplitude of damped oscillatory motion is exponential
Answer:
It can go back to it's original shape
Explanation:
Answer:
a) E = ρ / e0
b) E = ρ*a / (e0 * r)
c) E = 0
Explanation:
Because of the geometry, the electric field lines will all have a radial direction.
Using Gauss law

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.
We can make this cylinder of different radii to evaluate the electric field at different points.
Then:
A = 2*π*r (area of cylinder per unit of length)
Q/e0 = 2*π*r*E
E = Q / (2*π*e0*r)
Where Q is the charge contained inside the cylinder.
Inside the cable core:
There is a uniform charge density ρ
Q(r) = ρ * 2*π*r
Then
E = ρ * 2*π*r / (2*π*e0*r)
E = ρ / e0 (electric field is constant inside the charged cylinder.
Between ther inner cilinder and the tube:
Q = ρ * 2*π*a
E = ρ * 2*π*a / (2*π*e0*r)
E = ρ*a / (e0 * r)
Outside the tube, the charges of the core cancel each other.
E=0
This question is incomplete, the complete question is;
A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?
(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)
Answer: the maximum charge q is 716.85 μF
Explanation:
Given data;
with = 3.0 cm = 0.03
breathe = 5.0 m
Area = 0.03 × 5 = 0.15 m²
dielectric strength E = 1.00 × 10⁸
∈₀ = 8.85 × 10⁻¹²
constant K = 5.4
maximum charge = ?
the capacitor C = KA∈₀ / d
q = cv so c = q/v
now
q/v = KA∈₀ / d
q = vKA∈₀/d = EKA∈₀
we substitute
q = (1.00 × 10⁸) × 5.4 × 0.15 × 8.85 × 10⁻¹²
q = 716.85 × 10⁻⁶ F
q = 716.85 μF
the maximum charge q is 716.85 μF