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Show that the energy associated with a conducting sphere of radius \(R\) and charge \(Q\) surrounded by a vacuum is \(U=k_{e} Q^{2} / 2 R .\)

Short Answer

Expert verified
\(U = \frac{k_e Q^2}{2R}\)

Step by step solution

01

Consider a differential charge element

Introduce a differential charge element \(dq\) to the sphere's surface. The energy \(dU\) required to bring this charge from infinity to the sphere's surface is equal to the work done against the electrostatic force.
02

Calculate the potential at the sphere's surface

Use the formula for potential \( V = \frac{k_e Q}{R} \) to calculate the potential at the surface of the sphere, where \( k_e \) is Coulomb's constant, \( Q \) is the total charge on the sphere, and \( R \) is the radius of the sphere.
03

Relate differential work to the potential

The work done \(dU\) to bring the differential charge \(dq\) from infinity to the radius \(R\) can be expressed as \( dU = V dq = \frac{k_e Q}{R} dq \) since the voltage is constant all over the sphere's surface.
04

Integrate to find the total energy

Integrate the expression for \(dU\) from 0 to \(Q\) to find the total energy \(U\). The integral is \(\int_0^Q \frac{k_e Q}{R} dq = \frac{k_e}{R} \int_0^Q Q dq = \frac{k_e}{R} \frac{Q^2}{2},\) because the charge \(Q\) is built up gradually from 0 to the final charge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential
Electric potential is a fundamental concept in electrostatics that represents the potential energy per unit charge. Think of it as a measure of how much work would be required to bring a positive unit charge from a point at infinity to a specific point within an electric field without producing any acceleration. For a conducting sphere, the electric potential on the surface can be described by the formula:
\( V = \frac{k_e Q}{R} \),
where \(k_e\) is Coulomb's constant, \(Q\) is the total charge on the sphere, and \(R\) is the radius of the sphere. This potential is due to the electric field produced by the charge distributed over the sphere's surface. The electric potential is the same at every point on the surface because a conductor redistributes its charge until it reaches equilibrium, where the surface is an equipotential surface.
Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest. The forces between stationary charges are described by Coulomb's law, which states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Conductors are materials that allow charges to move freely. When dealing with a conducting sphere, charges reside on the surface, creating a uniform electric field just outside the surface and no electric field inside the sphere. This uniform distribution is key to understanding the behavior and calculations of potential and energy in electrostatic scenarios.
Work-Energy Theorem
The work-energy theorem is a fundamental principle that connects the work done on an object to its energy. In the context of electrostatics, the work done to move a charge in an electric field directly relates to the change in electric potential energy of the system. To move a small charge \(dq\) from infinity to the surface of a conducting sphere, the work done is equal to the increase in potential energy of the charge. This work is defined using the concept of the electric potential, and over the process of assembling the sphere's charge, the work done is accumulated. This accumulation is integral to understanding how the energy of the sphere is calculated.
Coulomb's Constant
Coulomb's constant \(k_e\) is a proportionality factor that appears in Coulomb's law and electric potential calculations. Its value is approximately \(8.9875 \times 10^9 N m^2 C^{-2}\) in the SI unit system. This constant is crucial for calculating the forces between charged particles and the electric potential due to a particular distribution of charges. In the formula for the energy associated with a conducting sphere \(U = \frac{k_e Q^2}{2R}\), Coulomb's constant links the geometrical and charge properties of the sphere to its electrostatic energy. Understanding \(k_e\) is essential for making precise calculations in electrostatics and comprehending the strength of electric interactions across distances.

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Most popular questions from this chapter

As a person moves about in a dry environment, electric charge accumulates on his body. Once it is at high voltage, either positive or negative, the body can discharge via sometimes noticeable sparks and shocks. Consider a human body well separated from ground, with the typical capacitance 150 \(\mathrm{pF}\) . (a) What charge on the body will produce a potential of 10.0 \(\mathrm{kV} ?\) (b) Sensitive electronic devices can be destroyed by electrostatic discharge from a person. A particular device can be destroyed by a discharge releasing an energy of 250\(\mu \mathrm{J}\) . To what voltage on the body does this correspond?

An air-filled spherical capacitor is constructed with inner and outer shell radii of 7.00 and \(14.0 \mathrm{cm},\) respectively. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a charge of 4.00\(\mu \mathrm{C}\) on the capacitor?

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) A sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform electric field at nearby points, with magnitude \(E=Q / 2 A \epsilon\) . (b) Two large sheets of area \(A,\) carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential \(Q d / A \epsilon\) (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d .\)

A 50.0 -m length of coaxial cable has an inner conductor that has a diameter of 2.58 \(\mathrm{mm}\) and carries a charge of \(8.10 \mu \mathrm{C} .\) The surrounding conductor has an inner diameter of 7.27 \(\mathrm{mm}\) and a charge of \(-8.10 \mu \mathrm{C} .\) (a) What is the capacitance of this cable? (b) What is the potential difference between the two conductors? Assume the region between the conductors is air.

An air-filled capacitor consists of two parallel plates, each with an area of \(7.60 \mathrm{cm}^{2},\) separated by a distance of \(1.80 \mathrm{mm} .\) A \(20.0-\mathrm{V}\) potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate.

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