For thorough details and necessary calculations, please refer to the attachment.
The infinitesimal charge dQ on a layer with thickness dr is expressed as
dQ = (charge density) × (surface area) × dr
dQ = ρ(r)4πr²dr
∫ dQ = ∫ (a/r)4πr²dr
∫ dQ = 4πa ∫ rdr
Q(r) = 2πar² - 2πa0²
Q = 2πar² (= total charge confined within a spherical surface of radius r)
According to Gauss's Law:
(Flux through surface) = (charge enclosed by surface)/ε۪
(Surface area of sphere) × E = Q/ε۪
4πr²E = 2πar²/ε۪
<span>E = a/2ε۪
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Speed is defined as distance over time. Hence, to determine the distance, we use d = V * t. Plugging in the values yields d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km. Thus, during this distracted period, a distance of 0.08Km was covered.
Answer: The result to the query is 0.25 ohms
Explanation:
R = u x/A.......1
where u represents the resistivity of the
rod, A is the cross-sectional area, and x denotes
the length of the rod.
Let R* represent the resistance across the adjacent sections of the rod.
Then, R* = u1/4.......2
By comparing equation 1 with equation 2, we find that
R* = 1/4
which equals 0.25 ohms.
Answer:
The electric flux going through the sphere is 
Explanation:
Given
Required
Calculate the electric flux
Electric flux can be computed using the formula;
Ф = q/ε
Where ε stands for the electric constant permittivity
ε 
Substituting ε and 
; the formula simplifies to
Ф = 
Ф = 
Ф = 
Ф = 
Ф = 
Ф = 
Ф = 
Ф =
Thus, the electric flux through the sphere is
