Question

Нужно решить задачу Электрическое поле создано бесконечной плоскостью с поверхностной плотностью заряда =200 нКл/ и бесконечной нитью с линейной плотностью =0,1 мкКл/м, проходящей параллельно плоскости на расстоянии а=0,2 м. Определить напряженность поля в точке на расстоянии =0,5 м от плоскости и =0,3 м от нити.

Answer 1

Ответ:

Е = 17 кВ/м

Объяснение:

Дано:

σ = 200 нКл/м² = 200·10⁻⁹ Кл/м²

τ = 0,1 мкКл/м = 0,1·10⁻⁶ Кл/м

a = 0,2 м

L = 0,5 м

r = 0,3 м

__________________

E - ?

1)

Сделаем чертеж.

Заметим, что напряженность поля, созданная заряженной плоскостью, не зависит от расстояния от плоскости до исследуемой точки:

E₁ = σ / (2·ε₀).

2)

Напряженность поля, созданная заряженной нитью, от расстояния зависит:

E₂ = τ / (2·π·ε₀·r).

3)

По принципу суперпозиции полей:

E = E₁+E₂

E = σ / (2·ε₀) + τ / (2·π·ε₀·r) = (1/(2·ε₀)) · (σ + τ / (π·r) )

E =  (1/(2·8,85·10⁻¹²)) · (200·10⁻⁹ + 0,1·10⁻⁶ / (3,14·0,3) ) =

= 5,65·10¹⁰ · (200·10⁻⁹ + 106·10⁻⁹) ≈  17 000 В/м   или   Е=17 кВ/м

Answer 2

Problem Statement

We are given an electric field created by an infinite plane with a surface charge density of 200 nC/m² and an infinite wire with a linear charge density of 0.1 μC/m, running parallel to the plane at a distance of a = 0.2 m. We need to determine the electric field intensity at a point located at a distance of 0.5 m from the plane and 0.3 m from the wire.

Solution

To solve this problem, we can break it down into two parts: the contribution of the plane and the contribution of the wire to the electric field at the given point.

# Electric Field Contribution from the Plane

The electric field intensity due to an infinite plane with a surface charge density can be calculated using the formula:

E_plane = σ / (2 * ε₀)

Where: - E_plane is the electric field intensity due to the plane, - σ is the surface charge density, and - ε₀ is the permittivity of free space.

Given that the surface charge density is 200 nC/m², we can substitute the values into the formula:

E_plane = (200 nC/m²) / (2 * ε₀)

Now, we need to find the value of the permittivity of free space, ε₀. The permittivity of free space is a fundamental constant with a value of approximately 8.854 x 10⁻¹² C²/(N·m²).

E_plane = (200 nC/m²) / (2 * 8.854 x 10⁻¹² C²/(N·m²))

Simplifying the expression:

E_plane = (200 x 10⁻⁹ C/m²) / (2 * 8.854 x 10⁻¹² C²/(N·m²))

E_plane = (200 x 10⁻⁹ C/m²) / (2 x 8.854 x 10⁻¹² C²/(N·m²))

E_plane = (200 x 10⁻⁹ C/m²) / (17.708 x 10⁻¹² C²/(N·m²))

E_plane = (200 / 17.708) x (10⁻⁹ / 10⁻¹²) N/C

E_plane = 11.31 x 10³ N/C

Therefore, the electric field intensity due to the plane at the given point is 11.31 x 10³ N/C.

# Electric Field Contribution from the Wire

The electric field intensity due to an infinite wire with a linear charge density can be calculated using the formula:

E_wire = (λ / (2 * π * ε₀ * r)) * ln(b / a)

Where: - E_wire is the electric field intensity due to the wire, - λ is the linear charge density, - ε₀ is the permittivity of free space, - r is the distance from the wire, - a is the distance from the wire to the point where the field is being calculated, and - b is the distance from the wire to a reference point.

Given that the linear charge density is 0.1 μC/m and the distances are as follows: - r = 0.3 m (distance from the wire to the point where the field is being calculated) - a = 0.2 m (distance from the wire to the plane) - b = 0.5 m (distance from the wire to a reference point)

We can substitute the values into the formula:

E_wire = (0.1 μC/m) / (2 * π * ε₀ * 0.3 m) * ln(0.5 m / 0.2 m)

Now, we need to find the value of the natural logarithm of the ratio of b/a. Using a calculator, we find that ln(0.5/0.2) ≈ 0.916.

E_wire = (0.1 x 10⁻⁶ C/m) / (2 * π * 8.854 x 10⁻¹² C²/(N·m²) * 0.3 m) * 0.916

E_wire = (0.1 x 10⁻⁶ C/m) / (2 * π * 8.854 x 10⁻¹² C²/(N·m²) * 0.3 m) * 0.916

E_wire = (0.1 / (2 * π * 8.854 * 0.3)) x (10⁻⁶ / 10⁻¹²) N/C

E_wire = (0.1 / (2 * π * 8.854 * 0.3)) x (10⁶ / 1) N/C

E_wire ≈ 18.96 x 10³ N/C

Therefore, the electric field intensity due to the wire at the given point is approximately 18.96 x 10³ N/C.

Final Result

The total electric field intensity at the given point, considering the contributions from both the plane and the wire, is the sum of the individual contributions:

E_total = E_plane + E_wire

E_total = 11.31 x 10³ N/C + 18.96 x 10³ N/C

E_total ≈ 30.27 x 10³ N/C

Therefore, the electric field intensity at the given point is approximately 30.27 x 10³ N/C.