Calculate The Electrostatic Potential Energy Of A Uniformly Charged Solid Sphere With Radius R And Charge Q.
Introduction
In physics, understanding the electrostatic potential energy of charge distributions is crucial for analyzing the behavior of electric fields and forces. This article delves into calculating the electrostatic potential energy of a uniformly charged solid sphere, a classic problem in electrostatics. We will explore the concepts and mathematical techniques required to arrive at the correct solution, providing a comprehensive explanation for students and enthusiasts alike. Specifically, we will derive the formula for the electrostatic potential energy, which is a measure of the energy required to assemble the charge distribution from infinity. By understanding this calculation, we gain insights into the fundamental principles governing electrostatic interactions. Let's embark on this journey to unravel the intricacies of electrostatics and solve this fascinating problem.
Understanding Electrostatic Potential Energy
To begin, let's define electrostatic potential energy. It is the energy required to bring a charge from infinity to a specific point in an electric field or, in this case, to assemble a charge distribution. This concept is vital for understanding how charges interact and the energy stored in their configuration. The potential energy arises from the work done against the electrostatic forces to bring the charges together. When we talk about a uniformly charged solid sphere, we imagine building the sphere by bringing infinitesimally small charges from infinity, adding them to the sphere until it reaches its final radius R and total charge Q. Each time a small charge is added, it experiences the electric potential created by the charges already present. The cumulative work done in this process is the electrostatic potential energy of the sphere.
The formula for electrostatic potential energy (U) is given by the integral of the potential (V) times the charge element (dq) over the entire charge distribution:
U = (1/2) ∫ V dq
This integral represents the summation of the potential energy contributions from each infinitesimal charge element. The factor of 1/2 is necessary because each pair of charges contributes to the potential energy, and without this factor, we would be counting each interaction twice. The potential V at any point is influenced by all other charges present in the system. Thus, to calculate the total electrostatic potential energy, we need to carefully consider the charge distribution and the resulting electric potential at various points within and on the surface of the sphere. In the subsequent sections, we will apply this fundamental principle to derive the electrostatic potential energy of a uniformly charged solid sphere, step by step.
Problem Statement: Calculating Electrostatic Potential Energy
Our primary goal is to calculate the electrostatic potential energy (U) of a uniformly charged solid sphere. The sphere has a radius (R) and a total charge (Q), which is uniformly distributed throughout its volume. This means that the charge density (ρ) is constant and can be expressed as:
ρ = Q / (4/3 πR³)
To find the total electrostatic potential energy, we will use the concept of building the sphere layer by layer, bringing infinitesimal charges from infinity to the sphere. At each step, we need to calculate the potential due to the existing charge and the work done to add the next infinitesimal charge. This process involves integrating the potential over the charge distribution. The electrostatic potential energy is a fundamental property of the sphere and provides insights into the stability and interactions of charged objects.
The problem can be mathematically represented as finding the energy required to assemble this charge distribution. We need to consider the electric potential created by the already assembled charge and the work done in bringing an additional charge element to the sphere. The solution involves integrating the potential energy contributions as we build the sphere from an initial state of no charge to its final state with charge Q and radius R. The challenge lies in setting up the integral correctly and performing the integration to obtain a closed-form expression for the electrostatic potential energy. The result will be expressed in terms of the given parameters k (Coulomb's constant), Q, and R, which will allow us to compare our solution with the provided options.
Methodology: Layer-by-Layer Construction
To calculate the electrostatic potential energy, we will use a layer-by-layer construction method. Imagine building the sphere by adding infinitesimally thin spherical shells of charge, one on top of the other, until we reach the final radius R. At each step, we calculate the potential due to the existing charge and the work done to add the next shell of charge. This method allows us to integrate the potential energy contribution incrementally, providing a systematic approach to the problem. The key is to express the charge of each shell (dq) in terms of the radius and the charge density (ρ), and then calculate the potential (V) at the surface of the sphere due to the existing charge.
The process can be described as follows:
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Consider a sphere of radius r (r < R) with charge q. The potential (V) at the surface of this sphere is given by:
V = kq/r
where k is Coulomb's constant.
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Now, add a thin spherical shell of charge dq to this sphere. The work done (dW) to bring this charge dq from infinity to the surface of the sphere is:
dW = V dq
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Express the charge dq of the thin shell in terms of the volume charge density (ρ) and the volume of the shell (4πr² dr):
dq = ρ dV = ρ (4πr² dr)
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Substitute the expression for ρ in terms of Q and R:
ρ = Q / (4/3 πR³) dq = (Q / (4/3 πR³)) (4πr² dr) = (3Q / R³) r² dr
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Substitute V and dq into the expression for dW:
dW = (kq/r) ((3Q / R³) r² dr)
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Express q in terms of r and ρ:
q = ρ (4/3 πr³) = (Q / (4/3 πR³)) (4/3 πr³) = (Qr³ / R³)
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Substitute q into the expression for dW:
dW = k (Qr³ / R³)/r (3Q / R³) r² dr = (3kQ² / R⁶) r⁴ dr
By integrating this expression for dW from 0 to R, we can find the total work done, which is equal to the electrostatic potential energy of the sphere. The next section will detail this integration process and derive the final formula.
Step-by-Step Calculation
Following the methodology described above, we now integrate the expression for dW to find the total electrostatic potential energy (U). We have:
dW = (3kQ² / R⁶) r⁴ dr
To find the total work done (U), we integrate dW from r = 0 to r = R:
U = ∫dW = ∫₀ᴿ (3kQ² / R⁶) r⁴ dr
Now, we perform the integration:
U = (3kQ² / R⁶) ∫₀ᴿ r⁴ dr U = (3kQ² / R⁶) [r⁵ / 5]₀ᴿ U = (3kQ² / R⁶) (R⁵ / 5) U = (3/5) (kQ² / R)
Therefore, the electrostatic potential energy of a uniformly charged solid sphere is:
U = (3/5) kQ²/R
This result shows that the electrostatic potential energy is directly proportional to the square of the charge (Q²) and inversely proportional to the radius (R) of the sphere. The constant of proportionality includes Coulomb's constant (k) and the factor 3/5, which arises from the uniform charge distribution within the sphere. This calculation demonstrates the interplay between charge, size, and energy in electrostatic systems.
Final Result and Answer
After performing the step-by-step calculation, we have arrived at the final result for the electrostatic potential energy (U) of a uniformly charged solid sphere with radius R and charge Q:
U = (3/5) kQ²/R
Comparing this result with the given options, we can see that it matches option (3):
(3) 3kQ²/5R
Therefore, the correct answer is (3). This result highlights the importance of understanding the concepts of electrostatic potential and energy, as well as the mathematical techniques involved in integrating charge distributions. The electrostatic potential energy of a charged sphere is a fundamental quantity in electromagnetism and has applications in various areas of physics.
Conclusion
In conclusion, we have successfully calculated the electrostatic potential energy of a uniformly charged solid sphere. By using the method of layer-by-layer construction and integrating the work done to bring infinitesimal charges from infinity, we derived the formula U = (3/5) kQ²/R. This formula provides a clear relationship between the charge, radius, and electrostatic potential energy of the sphere. The process involved understanding fundamental concepts such as electrostatic potential, charge density, and integration techniques.
This problem is a classic example in electrostatics and underscores the importance of careful, step-by-step calculations in physics. The correct answer, (3) 3kQ²/5R, demonstrates the precise balance between the charge and spatial distribution in determining the energy stored in an electric field. Understanding this calculation not only reinforces basic principles but also lays the groundwork for more complex problems in electromagnetism. The concepts and methods discussed in this article are crucial for anyone studying physics or electrical engineering and provide valuable insights into the behavior of charged systems.