ಠ_ಠ Hey, hang on.. you might've made a discovery. Nobody has tested it so how do we know? ಠ_ಠ
Answer:
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
Answer:
The linear charge density is 5.19 X 10⁻⁶ C/m
Explanation:
The potential difference between two cylinders, is given as
V = (λ/2πε)ln(b/a)
where;
λ is the line charge density on the power line.
b is the distance between the power line = 1 m
a is the radius of the wire = 1.5 cm = 0.015 m
ε is the permittivity of free space = 8.9 X 10⁻¹² C
V*2πε = λ* ln(b/a)
3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)
2.1812 X 10⁻⁷ = 4.1997* λ
λ = 5.19 X 10⁻⁶ C/m
Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m
It is commonly perceived as "thickness", or resistance to pouring. Viscosity describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction. Thus, water is "thin", having a low viscosity, while vegetable oil is "thick" having a high viscosity.
Answer:
8.8 × 10⁻³ g/L
Explanation:
NaF is a strong electrolyte that ionizes according to the following reaction.
NaF(aq) → Na⁺(aq) + F⁻(aq)
Then, the concentration of F⁻ will also be 0.10 M.
In order to find the solubility of PbF₂ (S), we will use an ICE Chart.
PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)
I 0 0.10
C +S +2S
E S 0.10 + 2S
The solubility product (Kps) is:
Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²
In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.
Kps = 3.6 × 10⁻⁸ = S . (0.10)²
S = 3.6 × 10⁻⁵ M
The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:
3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L
