The North Pole would be your answer
A bridge supported by vertical cables which then leads to more support from larger cables.
Answer:
Explanation:
If a number of less than 1, then the number has a decimal point like
0.085, 0.008 e.t.c.
The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.
But if there a zero between the none zero e.g. 0.0087056
Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6
But if the zero is not in between the none zero digit, then the zero is insignificant
E.g 0.05800
The last two zero is insignificant, the significant number is 5 and 8
So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.
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
This what i found hope it helps