Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = 
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
Answer:
a and c are both correct
Explanation:
The Greek Philosophers had to reject many existing explanations (mostly from the babylonians and the egyptians) because of possible religion bias. <u>They resorted to logic and reason to provide understanding for the world around them</u>. The first set of Greek Philosophers were the Milesian thinkers; <u>whom aside from logic and reason also placed value on Observation in understanding the world</u>.
There are generally no evidence to support the use of experiment to support the claims of the Greek philosophers
Molecular formulas:
- CH₂O;
- C₂H₄O₂;
- C₆H₁₂O₆.
<h3>Explanation</h3>
The empirical formula of a compound tells only the ratio between atoms of each element. The empirical formula CH₂O indicates that in this compound,
- for each C atom, there are
- two H atoms, and
- one O atom.
The molecular weight (molar mass) of the molecule depends on how many such sets of atoms in each molecule. The empirical formula doesn't tell anything about that number.
It's possible to <em>add</em> more of those sets of atoms to a molecular formula to increase its molar mass. For every extra set of those atoms added, the molar mass increase by the mass of that set of atoms. The mass of one mole of C atoms, two mole of H atoms, and one mole of O atoms is 
.
- CH₂O- 30.0 g/mol;
- C₂H₄O₂- 30.0 + 30.0 = 2 × 30.0 = 60.0 g/mol;
- C₃H₆O₃- 30.0 + 30.0 + 30.0 = 3 × 30.0 = 90.0 g/mol.
It takes one set of those atoms to achieve a molar mass of 30.0 g/mol. Hence the molecular formula CH₂O.
It takes two sets of those atoms to achieve a molar mass of 60.0 g/mol. Hence the molecular formula C₂H₄O₂.
It takes 
sets of those atoms to achieve a molar mass of 180.0 g/mol. Hence the molecular formula C₆H₁₂O₆.
