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Chapter 24: Problem 31

Consider a thin spherical shell of radius 14.0 \(\mathrm{cm}\) with a total charge of 32.0\(\mu \mathrm{C}\) distributed uniformly on its surface. Find the electric field (a) 10.0 \(\mathrm{cm}\) and \((\mathrm{b}) 20.0 \mathrm{cm}\) from the center of the charge distribution.

Short Answer

Expert verified
The electric field is 0 \(\frac{N}{C}\) at 10.0 \( \mathrm{cm} \) from the center (inside the shell) and 7192 \( \frac{N}{C} \) at 20.0 \( \mathrm{cm} \) from the center (outside the shell).

Step by step solution

01

Understand the Concept

The electric field due to a uniformly charged spherical shell is different depending on whether the point where we are calculating the field is outside the shell, inside the shell, or on the shell. For points outside, the shell can be considered as a point charge, with electric field given by Coulomb's law. Inside the shell, the field is zero.
02

Calculate the Electric Field at 10.0 cm (Inside the Shell)

Since the point at 10.0 cm is inside the spherical shell, the electric field is zero, because the shell's charge uniformly distributes the field.
03

Convert Charge to Coulombs

First, convert the charge from microcoulombs to coulombs by using the conversion factor 1 \( \mu C = 10^{-6} C \). \[ Q = 32.0 \mu C = 32.0 \times 10^{-6} C \]
04

Apply Coulomb's Law for the Point at 20.0 cm (Outside the Shell)

Use Coulomb's law to calculate the electric field at a point outside the spherical shell. The formula for the electric field \(E\) at a distance \(r\) from a point charge \(Q\) is \[ E = \frac{k \cdot Q}{r^2}, \] where \(k\) is Coulomb's constant \(k = 8.99 \times 10^9 \frac{N \cdot m^2}{C^2}\).
05

Calculate the Electric Field at 20.0 cm

Use the given radius \(r = 20.0 \mathrm{cm} = 0.20 \mathrm{m}\) and the converted charge \(Q = 32.0 \times 10^{-6} C\) to calculate the electric field. \[ E = \frac{(8.99 \times 10^9) \cdot (32.0 \times 10^{-6})}{(0.20)^2} \]
06

Simplify the Calculation

Carry out the necessary mathematical operations to find the electric field value: \[ E = \frac{(8.99 \times 10^9) \cdot (32.0 \times 10^{-6})}{0.04} \] \[ E = \frac{287.68}{0.04} \] \[ E = 7192 \frac{N}{C} \]

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle that describes the force between two point charges. According to the law, the electrostatic force (\f\( F \f)\) between two charges is directly proportional to the product of the charges (\f\( q1 \f\) and \f\( q2 \f\)) and inversely proportional to the square of the distance (\f\( r \f\)) between their centers.

The mathematical formula for Coulomb's Law is given by:
\f[ F = k \frac{|q1 \times q2|}{r^2} \f]
where \f\( k \f\) is the Coulomb's constant, approximately \f\( 8.99 \times 10^9 \frac{N \times m^2}{C^2} \f\). This law is essential for understanding the electric field generated by a point charge, and it applies perfectly to a spherical shell when considering points outside the shell, treating it as if all its charge were concentrated at its center.
Electric Charge Distribution
The distribution of electric charge on a conductor or insulator defines the resulting electric field. For a spherical shell with a uniform charge distribution, each little charge element creates a contribution to the electric field.

When the point of interest is inside such a shell, as stated in the exercise solution, the contributions from all these charge elements exactly cancel each other out due to their symmetrical arrangement. That's why the electric field inside a uniformly charged spherical shell is zero, illustrating an intriguing property of electric field symmetry and distribution.
Electric Field Calculations
For calculating electric fields, we use both the geometry of the charge distribution and Coulomb's Law. In the case of a point outside a uniformly charged spherical shell, as shown in the exercise, the shell behaves as if all its charge were located at the center.

The electric field (\f\( E \f\)) at a point in space due to a charge (\f\( Q \f\)) at a distance (\f\( r \f\)) is given by the equation:
\f[ E = \frac{k \times Q}{r^2} \f]
This equation is derived from Coulomb's Law and allows us to calculate the magnitude of the electric field at a given point. For example, to find the electric field 20.0 cm from the center of the charge distribution in the given problem, we substitute the relevant values into the formula, resulting in an electric field strength of \f\( 7192 \frac{N}{C} \f\). Understanding this formula is key to solving many problems related to the electric field.

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Most popular questions from this chapter

A conducting spherical shell of inner radius \(a\) and outer radius \(b\) carries a net charge \(Q\) . A point charge \(q\) is placed at the center of this shell. Determine the surface charge density on (a) the inner surface of the shell and (b) the outer surface of the shell.

An early (incorrect) model of the hydrogen atom, suggested by J.J. Thomson, proposed that a positive cloud of charge \(+e\) was uniformly distributed throughout the volume of a sphere of radius \(R,\) with the electron an equal- magnitude negative point charge \(-e\) at the center. (a) Using Gauss's law, show that the electron would be in equilibrium at the center and, if displaced from the center a distance \(r< R,\) would experience a restoring force of the form \(F=-K r,\) where \(K\) is a constant. (b) Show that \(K=k_{e} e^{2} / R^{3} .\) (c) Find an expression for the frequency \(f\) of simple harmonic oscillations that an electron of mass \(m_{e}\) would undergo if displaced a small distance \((< R)\) from the center and released. (d) Calculate a numerical value for \(R\) that would result in a frequency of \(2.47 \times 10^{15} \mathrm{Hz}\) , the frequency of the light radiated in the most intense line in the hydrogen spectrum.

A conducting spherical shell of radius 15.0 \(\mathrm{cm}\) carries a net charge of \(-6.40 \mu \mathrm{C}\) uniformly distributed on its surface. Find the electric field at points (a) just outside the shell and (b) inside the shell.

A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of \(\lambda\) , and the cylinder has a net charge per unit length of 2\(\lambda\) . From this information, use Gauss's law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance \(r\) from the axis.

An insulating solid sphere of radius \(a\) has a uniform volume charge density and carries a total positive charge Q. A spherical gaussian surface of radius \(r,\) which shares a common center with the insulating sphere, is inflated starting from \(r=0 .\) (a) Find an expression for the electric flux passing through the surface of the gaussian sphere as a function of \(r\) for \(r< a\) . (b) Find an expression for the electric flux for \(r>a\) (c) Plot the flux versus \(r .\)
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