Question

Two very large, nonconducting plastic sheets, each 10.0 cm

thick, carry uniform charge densities σ1,σ2,σ3
and σ4
on their surfaces, as shown in the following figure(Figure 1). These surface charge densities have the values σ1 = -7.30 μC/m2 , σ2=5.00μC/m2, σ3= 1.90 μC/m2 , and σ4=4.00μC/m2. Use Gauss's law to find the magnitude and direction of the electric field at the following points, far from the edges of these sheets.

A:What is the magnitude of the electric field at point A , 5.00 cm
from the left face of the left-hand sheet?(Express your answer with the appropriate units.)

B:What is the direction of the electric field at point A, 5.00 cm
from the left face of the left-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)

C:What is the magnitude of the electric field at point B, 1.25 cm
from the inner surface of the right-hand sheet?(Express your answer with the appropriate units.)

D:What is the direction of the electric field atpoint B, 1.25 cm
from the inner surface of the right-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)

E:What is the magnitude of the electric field at point C , in the middle of the right-hand sheet?(Express your answer with the appropriate units.)

F:What is the direction of the electric field at point C, in the middle of the right-hand sheet?(LEFT,RIGHT,UPWARDS,DOWNWARDS)

Answer 1

Step-by-step explanation:

To use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.

Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.

Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.

A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively. Since the electric field is perpendicular to the faces, the flux through them is zero. So, Q_in = (σ1 - σ2) * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C

The magnitude of the electric field at point A is 2.31 x 10^5 N/C.

B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.

The direction of the electric field at point A is RIGHT.

C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C

The magnitude of the electric field at point B is 3.77 x 10^7 N/C.

D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.

The direction of the electric field at point B is LEFT.

E:

Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q

The net charge enclosed within this Gaussian surface is:

Q = σ1 × (2πrh)

where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:

Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C

Using Gauss's law, we can find the electric field at point C:

E × (2πrh) = Q/ε0

where ε0 is the permittivity of free space. Solving for E, we get:

E = Q / (2πε0rh)

Plugging in the values, we get:

E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C

Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.

To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.

The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:

qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)

where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.

Substituting the given values, we get:

qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC

Using Gauss's law, we have:

E * 2πrh = qenc/ε0

where ε0 is the permittivity of free space.

Solving for E, we get:

E = qenc / (2πrhε0) = 2.22 × 10^4 N/C

Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.

F:

The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.

Answer 2

A. The total electric field at point A is approximately 82.6 N/C, directed to the right.

B. The electric field at point A is directed **RIGHT**.

C. The total electric field at point B is approximately 45.2 N/C, directed to the left.

D. The electric field at point B is directed **LEFT**.

E. The total electric field at point C is approximately **0 N/C**.

F. Since the electric field at point C is zero, it has no direction. We can say it points **NOWHERE**.

Solving for the Electric Field using Gauss's Law:

Sure, I can help you solve this problem using Gauss's Law! Here's how we can find the magnitude and direction of the electric field at points A, B, and C:

**Gauss's Law:**

Gauss's Law states that the electric flux through any closed surface is proportional to the net charge enclosed by that surface. Mathematically, it can be expressed as:

Φ_E = ∑_i Q_i / ε_0

where:

* Φ_E is the electric flux through the surface

* Q_i is the net charge enclosed by the surface

* ε_0 is the permittivity of free space (8.854 × 10⁻¹² C²/N·m²)

**Applying Gauss's Law to each point:**

**A. Electric Field at Point A:**

* Choose a Gaussian surface in the form of a rectangular prism enclosing point A, with one face coinciding with the left face of the left-hand sheet.

* Since the sheet is non-conducting and very large, the net charge enclosed by the surface is only due to the leftmost sheet (σ₁).

* Applying Gauss's Law, the electric field (E) on the left face of the prism (due to σ₁) is:

E₁ = σ₁ / ε₀ = -7.30 × 10⁻⁶ C/m² / 8.854 × 10⁻¹² C²/N·m² ≈ 82.6 N/C (directed towards the right)

* Since the other faces of the prism are parallel to the sheet with no net charge, the electric field contribution from those faces is zero.

* Therefore, the total electric field at point A is approximately 82.6 N/C, directed to the right.

**B. Direction of the Electric Field at Point A:**

* As determined in part A, the electric field at point A is directed **RIGHT**.

**C. Electric Field at Point B:**

* Choose a Gaussian surface in the form of a rectangular prism enclosing point B, with one face coinciding with the inner surface of the right-hand sheet.

* The net charge enclosed by the surface is due to both the rightmost sheet (σ₄) and the inner portion of the right-hand sheet (σ₂ + σ₃).

* Applying Gauss's Law, the electric field (E) on the inner face of the prism (due to σ₄) is:

E₄ = σ₄ / ε₀ = 4.00 × 10⁻⁶ C/m² / 8.854 × 10⁻¹² C²/N·m² ≈ 45.2 N/C (directed towards the left)

* The electric field contribution from the other faces depends on the net charge distribution within the right-hand sheet. However, since the sheet is very large and σ₂ and σ₃ are uniform, their contributions to the electric field at point B cancel out.

* Therefore, the total electric field at point B is approximately 45.2 N/C, directed to the left.

**D. Direction of the Electric Field at Point B:**

* As determined in part C, the electric field at point B is directed **LEFT**.

**E. Electric Field at Point C:**

* Choose a Gaussian surface in the form of a rectangular prism enclosing point C, with one face coinciding with the middle of the right-hand sheet.

* The net charge enclosed by the surface is due to both halves of the right-hand sheet (σ₂ and σ₃).

* Applying Gauss's Law, the electric field contributions from both faces cancel out due to the equal and opposite charges from σ₂ and σ₃.

* Therefore, the total electric field at point C is approximately **0 N/C**.

**F. Direction of the Electric Field at Point C:**

* Since the electric field at point C is zero, it has no direction. We can say it points **NOWHERE**.