/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A charged isolated metal sphere ... [FREE SOLUTION] | 91Ó°ÊÓ

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A charged isolated metal sphere of diameter \(10 \mathrm{~cm}\) has a potential of \(8000 \mathrm{~V}\) relative to \(V=0\) at infinity. Calculate the energy density in the electric field near the surface of the sphere.

Short Answer

Expert verified
The energy density near the sphere's surface is calculated using the electric field derived from its potential.

Step by step solution

01

Finding the Radius

The sphere's diameter is given as 10 cm, so its radius \( r \) is half of that. Thus, \( r = \frac{10}{2} = 5 \) cm or \( 0.05 \) m.
02

Understanding Potential and Charge

For a sphere, the potential \( V \) at its surface is given by \( V = \frac{kQ}{r} \), where \( Q \) is the charge, \( r \) is the radius, and \( k \) is Coulomb's constant \( 8.99 \times 10^{9} \mathrm{~Nm}^{2}/ ext{C}^{2} \). We solve for \( Q \) using \( Q = \frac{Vr}{k} \).
03

Substitute values to find Charge

Substituting the known values: \( V = 8000 \mathrm{~V} \), \( r = 0.05 \) m, and \( k = 8.99 \times 10^{9} \mathrm{~Nm}^{2}/ ext{C}^{2} \), solve for \( Q \):\[Q = \frac{8000 \times 0.05}{8.99 \times 10^{9}}\]Calculate \( Q \).
04

Calculate Electric Field Near the Surface

Using the formula for the electric field \( E \) just outside the surface, \( E = \frac{Q}{4\pi\varepsilon_0 r^2} \) where \( \varepsilon_0 \approx 8.85 \times 10^{-12} \mathrm{~C}^{2}/( ext{N} ext{m}^{2}) \), substitute the previously calculated charge \( Q \), the known radius \( r \), and \( \varepsilon_0 \). Calculate \( E\).
05

Calculate Energy Density

The energy density \( u \) of an electric field is given by \( u = \frac{1}{2} \varepsilon_0 E^2 \). Use the value of \( E \) found in the previous step to substitute into this formula and find \( u \).

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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 measure of the potential energy per unit charge at a point in an electric field. It helps us understand how much work is needed to move a charge within the field. In this problem, the electric potential is given relative to a reference point, which is set at infinity where the potential energy is zero. The given sphere has an electric potential of 8000 V, which means each unit charge at the sphere's surface has 8000 joules of potential energy.

To calculate the charge on the sphere, we use the formula for potential, \( V = \frac{kQ}{r} \), where \( V \) is the potential, \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the radius of the sphere. This equation connects the potential to the physical properties of the sphere and allows us to find the charge required to achieve a given potential.
  • Potential stories the energy configuration of charges
  • Measured in volts (V)
  • Higher the potential, higher the tendency to move charges
Energy Density
Energy density in an electric field measures how much energy is stored in a given volume of space. It is important because it gives insights into how effectively an electric field can perform work when influencing charges. The energy density \( u \) is calculated using the formula \( u = \frac{1}{2} \varepsilon_0 E^2 \), where \( \varepsilon_0 \) is the permittivity of free space and \( E \) is the electric field strength.

This approach provides the energy stored per cubic meter due to the electric field. High energy density means more energy is available over a smaller area which can be critical in fields like capacitors where efficiency is key. In this problem, calculating \( E \) gives us the necessary information to determine the energy density at the sphere's surface.
  • Represents energy stored per unit volume
  • Significant for understanding electric fields in capacitors
  • Higher fields lead to greater energy density
Coulomb's Law
Coulomb's Law provides a quantitative description of the force between two charged objects. This fundamental principle is key for understanding electric forces at a distance, and it is defined as \( F = \frac{k |Q_1Q_2|}{r^2} \), where \( F \) is the force between charges, \( k \) is Coulomb's constant, \( Q_1 \) and \( Q_2 \) are the charges, and \( r \) is the distance between them.

In the context of this exercise, Coulomb's Law underlies the formula for electric potential and forms the foundation for understanding how charges interact in space. It demonstrates that the force is inversely proportional to the square of the distance between them, illustrating that forces diminish swiftly with distance.
  • Explains electric force between charged bodies
  • Directly proportional to charge magnitudes
  • Inversely proportional to the square of the distance apart
Electric Field
The electric field is a vector field that represents the force exerted by a charged object on other charges in its vicinity. It is a vital concept in electrostatics, defining how charges influence one another through space. The electric field \( E \) at the surface of a charged sphere can be calculated using the formula \( E = \frac{Q}{4\pi\varepsilon_0 r^2} \).

Electric fields detail both the direction and magnitude of the force acting on a positive charge. They indicate that the force is strongest nearest the charge and weaker with increased distance. In this exercise, once \( Q \) is known, calculating the electric field gives us relevant values for energy density analysis around the sphere. The formula incorporates properties of the sphere and constants like the permittivity of free space.
  • Vector quantity showing force influence
  • Strength and direction defined per unit charge
  • Reduces with distance from the source charge

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Most popular questions from this chapter

A certain parallel-plate capacitor is filled with a dielectric for which \(\kappa=5.5 .\) The area of each plate is \(0.034 \mathrm{~m}^{2},\) and the plates are separated by \(2.0 \mathrm{~mm} .\) The capacitor will fail (short out and burn up) if the electric field between the plates exceeds \(200 \mathrm{kN} / \mathrm{C}\). What is the maximum energy that can be stored in the capacitor?

Two parallel-plate capacitors, \(6.0 \mu \mathrm{F}\) each, are connected in parallel to a \(10 \mathrm{~V}\) battery. One of the capacitors is then squeezed so that its plate separation is \(50.0 \%\) of its initial value. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?

A \(10 \mathrm{~V}\) battery is connected to a series of \(n\) capacitors, each of capacitance \(2.0 \mu \mathrm{F}\). If the total stored energy is \(25 \mu \mathrm{J},\) what is \(n ?\)

The space between two concentric conducting spherical shells of radii \(b=1.70 \mathrm{~cm}\) and \(a=1.20 \mathrm{~cm}\) is filled with a substance of dielectric constant \(\kappa=23.5 .\) A potential difference \(V=73.0 \mathrm{~V}\) is applied across the inner and outer shells. Determine (a) the capacitance of the device, (b) the free charge \(q\) on the inner shell, and (c) the charge \(q^{\prime}\) induced along the surface of the inner shell.

You have two flat metal plates, each of area \(1.00 \mathrm{~m}^{2},\) with which to construct a parallel-plate capacitor. (a) If the capacitance of the device is to be \(1.00 \mathrm{~F},\) what must be the separation between the plates? (b) Could this capacitor actually be constructed?

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