Answer:
Explanation:
Given that
Emissivity of surfaces(∈) = 1
We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies
So now by putting the values
So rate of heat transfer per unit area
Charge q is 1 unit of distance away from the source charge S. Charge p is four times further away. The force exerted between S a
2 answers:
The correct answer is:
sixteen times
In fact, the distance between charge q and the source S is 1 unit. Instead, the distance between charge p and the source S is 4 units. The magnitude of the electrostatic force is inversely proportional to the square of the distance between the charge and the source:
where r is the distance. If we take the force between q and S as reference, we have r=1, so that
while the force between p and S is
Therefore, we see that the force exerted between q and S is 16 times the force exerted between p and S.
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Explanation:
Given that,
Charge on a spherical drop of water is 43 pC
The potential at its surface is 540 V
(a) The electric potential on the surface is given by :
r is the radius of the drop
(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,
Now the charge on the new drop is 2q. New potential is given by :
Hence, the radius of the drop is and the potential at the surface of the new drop is 856.79 V.
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
