An electric eel develops a potential difference of 470 V , driving a current of 0.85 A for a 1.0 ms pulse. Part A Find the power
1 answer:
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The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
<h3>What is an electric field?</h3>An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
The given data in the problem is given by;
E is the electric field = (200 N/C)
d is the distance = 5.0 cm.=0.05 m
Q is the charge of electrons= 1.602 x 10^-19 C
The formula for electric potential is given by;
The work is defined as the product of the potential difference and charge of an electron.
Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
To learn more about the electric field refer to the link;
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
