A Positive Charge Is Located Above The Gaussian Cylinder

A positive charge hovering above a Gaussian cylinder presents a fascinating problem in electrostatics. The electric field generated by this charge permeates the cylinder, leading to a non-uniform flux through its surfaces. Understanding this flux, and how it relates to the enclosed charge (or lack thereof), is crucial for mastering Gauss's Law and its applications. This article will delve into the intricacies of this scenario, exploring the concepts, calculations, and implications involved.

Introduction to Gauss's Law and Gaussian Surfaces

At the heart of this problem lies Gauss's Law, one of the fundamental principles of electromagnetism. Gauss's Law states that the total electric flux through any closed surface is proportional to the enclosed electric charge. Mathematically, it's expressed as:

∮ E ⋅ dA = Qenc / ε₀

Where:

• ∮ E ⋅ dA represents the electric flux through the closed surface.
• E is the electric field vector.
• dA is an infinitesimal area vector pointing outward from the surface.
• Qenc is the net electric charge enclosed by the surface.
• ε₀ is the permittivity of free space (approximately 8.854 × 10⁻¹² C²/Nm²).

Gaussian surfaces are hypothetical closed surfaces used to apply Gauss's Law. The choice of a Gaussian surface is strategic; ideally, it should be chosen such that the electric field is either constant and perpendicular to the surface or parallel to the surface (meaning the flux is zero). This simplifies the calculation of the flux integral. Common Gaussian surfaces include spheres, cylinders, and cubes, chosen based on the symmetry of the charge distribution.

In our case, we're dealing with a Gaussian cylinder and a point charge positioned above it. The lack of symmetry introduced by the off-center charge makes the problem more interesting.

Setting Up the Problem: A Positive Charge Above a Gaussian Cylinder

Imagine a cylindrical surface of radius r and height h. A positive point charge, +q, is located a distance d directly above the center of the top surface of the cylinder. Our objective is to determine the electric flux through the entire cylindrical surface and analyze its implications.

To approach this problem effectively, we need to consider the following:

• The Electric Field: The positive charge +q creates an electric field that radiates outward in all directions. The strength of the electric field at any point is given by Coulomb's Law:

  E = k * q / R²

  Where:

  • k is Coulomb's constant (approximately 8.9875 × 10⁹ Nm²/C²).
  • R is the distance from the charge +q to the point in question.

• The Cylindrical Surface: The cylinder consists of three distinct surfaces:

  • Top Surface: A circle of radius r at the top of the cylinder.
  • Bottom Surface: A circle of radius r at the bottom of the cylinder.
  • Curved Surface: The cylindrical wall connecting the top and bottom surfaces.

• Flux Calculation: The total flux through the cylinder is the sum of the flux through each of these three surfaces:

  Φtotal = Φtop + Φbottom + Φcurved

Calculating the Electric Flux Through Each Surface

The key to solving this problem lies in calculating the electric flux through each of the three surfaces of the cylinder. This involves evaluating the surface integral ∮ E ⋅ dA for each surface. Due to the non-uniformity of the electric field, this calculation can be complex.

1. Flux Through the Top Surface (Φtop)

The electric field lines emanating from the positive charge pass through the top surface of the cylinder. The angle between the electric field vector E and the area vector dA (which points outward, away from the cylinder) varies across the surface. Therefore, we need to integrate.

Consider a small area element dA on the top surface. The distance R from the charge +q to this area element is given by:

R = √(d² + ρ²)

Where ρ is the radial distance from the center of the top surface to the area element dA (ρ varies from 0 to r).

The angle θ between the electric field E and the area vector dA is given by:

cos θ = d / R = d / √(d² + ρ²)

The flux through the small area element dA is:

dΦtop = E ⋅ dA = E * dA * cos θ = (k * q / R²) * dA * (d / R) = (k * q * d * dA) / (d² + ρ²)^(3/2)

To find the total flux through the top surface, we integrate over the entire surface:

Φtop = ∫ dΦtop = ∫ (k * q * d * dA) / (d² + ρ²)^(3/2)

Since dA = ρ dρ dφ (in polar coordinates), the integral becomes:

Φtop = k * q * d ∫(φ=0 to 2π) ∫(ρ=0 to r) (ρ dρ dφ) / (d² + ρ²)^(3/2)

The integral over φ is simply 2π. Let u = d² + ρ², then du = 2ρ dρ. The integral over ρ becomes:

∫ (ρ dρ) / (d² + ρ²)^(3/2) = (1/2) ∫ du / u^(3/2) = -u^(-1/2) = -1/√(d² + ρ²)

Evaluating this from ρ = 0 to ρ = r, we get:

-1/√(d² + r²) - (-1/√(d² + 0²)) = 1/d - 1/√(d² + r²)

Therefore,

Φtop = k * q * d * 2π * [1/d - 1/√(d² + r²)] = 2πkq [1 - d/√(d² + r²)]

2. Flux Through the Bottom Surface (Φbottom)

Similar to the top surface, the electric field lines also pass through the bottom surface. However, the angle between the electric field vector E and the area vector dA (which points outward, away from the cylinder) is different. Furthermore, the distance from the charge to the bottom surface is now √( (d+h)² + ρ² ).

Following a similar procedure as for the top surface, we can derive the flux through the bottom surface:

The distance R from the charge +q to the small area element dA on the bottom surface is given by:

R = √((d + h)² + ρ²)

The angle θ between the electric field E and the area vector dA is such that cos θ = (d+h) / R = (d+h) / √((d+h)² + ρ²). Note that the area vector dA on the bottom surface points downwards (outwards from the closed surface), while the z-component of the electric field also points downwards.

The flux through the small area element dA is:

dΦbottom = E ⋅ dA = E * dA * cos θ = (k * q / R²) * dA * ((d+h) / R) = (k * q * (d+h) * dA) / ((d+h)² + ρ²)^(3/2)

To find the total flux through the bottom surface, we integrate over the entire surface:

Φbottom = ∫ dΦbottom = ∫ (k * q * (d+h) * dA) / ((d+h)² + ρ²)^(3/2)

Since dA = ρ dρ dφ (in polar coordinates), the integral becomes:

Φbottom = k * q * (d+h) ∫(φ=0 to 2π) ∫(ρ=0 to r) (ρ dρ dφ) / ((d+h)² + ρ²)^(3/2)

The integral over φ is simply 2π. Let u = (d+h)² + ρ², then du = 2ρ dρ. The integral over ρ becomes:

∫ (ρ dρ) / ((d+h)² + ρ²)^(3/2) = (1/2) ∫ du / u^(3/2) = -u^(-1/2) = -1/√((d+h)² + ρ²)

Evaluating this from ρ = 0 to ρ = r, we get:

-1/√((d+h)² + r²) - (-1/√((d+h)² + 0²)) = 1/(d+h) - 1/√((d+h)² + r²)

Therefore,

Φbottom = k * q * (d+h) * 2π * [1/(d+h) - 1/√((d+h)² + r²)] = 2πkq [1 - (d+h)/√((d+h)² + r²)]

3. Flux Through the Curved Surface (Φcurved)

The flux through the curved surface is the most challenging to calculate directly. The electric field E varies in both magnitude and direction along the surface. The angle between E and the area vector dA also changes continuously.

However, we can avoid a complex integration by using Gauss's Law itself. We know that the total flux through the entire closed surface must be zero because the charge +q is located outside the cylinder. Therefore,

Φtotal = Φtop + Φbottom + Φcurved = 0

Hence,

Φcurved = - (Φtop + Φbottom)

Φcurved = -2πkq [1 - d/√(d² + r²)] - 2πkq [1 - (d+h)/√((d+h)² + r²)]

Φcurved = -2πkq [2 - d/√(d² + r²) - (d+h)/√((d+h)² + r²)]

Total Flux and Verification of Gauss's Law

As we stated above, the total flux through the Gaussian cylinder must be zero because the positive charge +q is located outside the cylinder. Let's verify this by summing the flux through each surface:

Φtotal = Φtop + Φbottom + Φcurved

Φtotal = 2πkq [1 - d/√(d² + r²)] + 2πkq [1 - (d+h)/√((d+h)² + r²)] -2πkq [2 - d/√(d² + r²) - (d+h)/√((d+h)² + r²)]

Φtotal = 2πkq [1 - d/√(d² + r²) + 1 - (d+h)/√((d+h)² + r²) - 2 + d/√(d² + r²) + (d+h)/√((d+h)² + r²)]

Φtotal = 2πkq [0] = 0

This result confirms Gauss's Law: since no charge is enclosed within the Gaussian cylinder, the total electric flux through it is zero.

Implications and Key Takeaways

This problem, while seemingly simple, highlights several important aspects of Gauss's Law and electromagnetism:

• Gauss's Law and Enclosed Charge: The total electric flux through a closed surface depends only on the amount of charge enclosed by the surface, not on the location of the charge outside the surface.
• Strategic Choice of Gaussian Surface: While Gauss's Law is always true, its application is most effective when the Gaussian surface is chosen to exploit symmetry in the charge distribution. In this case, the lack of symmetry made direct calculation of the flux more challenging, but still possible.
• Superposition Principle: The electric field at any point is the vector sum of the electric fields due to all individual charges.
• Non-Uniform Electric Fields: Dealing with non-uniform electric fields often requires integration to calculate the electric flux accurately.
• Verification of Theoretical Results: It's always a good practice to verify theoretical results against known laws, like Gauss's Law, to ensure consistency and accuracy.

FAQ: Understanding the Nuances

Q: What if the positive charge was inside the cylinder?

A: If the charge was inside the cylinder, the total flux would not be zero. According to Gauss's Law, the total flux would be equal to q/ε₀. The distribution of the flux among the top, bottom, and curved surfaces would change, but the total flux would be determined solely by the enclosed charge.

Q: Would the result change if the cylinder was replaced by a cube or a sphere?

A: The total flux would still be zero if the charge remained outside the closed surface (cube or sphere). However, the distribution of the flux across the different surfaces of the cube or sphere would be different due to the different geometry. The calculations would also become more complex.

Q: Can we always use Gauss's Law to find the electric field?

A: While Gauss's Law is always valid, it's most useful for finding the electric field in situations with high symmetry (spherical, cylindrical, or planar symmetry). In these cases, we can choose a Gaussian surface such that the electric field is constant and perpendicular to the surface, simplifying the calculation. In cases with low symmetry, like our example, calculating the electric field directly using Coulomb's Law and superposition is often more practical. However, Gauss's Law is still invaluable for verifying the results.

Q: What if the charge was negative instead of positive?

A: If the charge was negative, the electric field would point inward towards the charge. The direction of the flux through each surface would be reversed, and the flux values (Φtop, Φbottom, Φcurved) would have opposite signs. However, the total flux would still be zero because the charge is outside the cylinder.

Conclusion

The problem of a positive charge located above a Gaussian cylinder provides a powerful illustration of Gauss's Law and its implications. By carefully calculating the electric flux through each surface of the cylinder and verifying the result against Gauss's Law, we gain a deeper understanding of electrostatics and the relationship between electric fields, charges, and closed surfaces. While the non-symmetric geometry makes the calculations more involved, it reinforces the fundamental principle that the total electric flux through a closed surface is solely determined by the enclosed charge. This problem serves as a valuable exercise for students and anyone seeking a more profound grasp of electromagnetism.