/*! 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 51 Negative charge \(-Q\) is distri... [FREE SOLUTION] | 91Ó°ÊÓ

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Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius \(R .\) Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located (a) a distance \(r > R\) from the center of the shell (outside the shell) and (b) a distance \(r < R\) from the center of the shell (inside the shell).

Short Answer

Expert verified
(a) Force is \( \frac{-Qq}{4\pi\varepsilon_0 r^2} \) inward; (b) Force is 0 inside.

Step by step solution

01

Applying Gauss's Law for Outside the Shell

According to Gauss's law, the electric field due to a spherical charge distribution outside the charge distribution is as if all the charge were concentrated at the center. The total charge on the shell is \( -Q \). A positive point charge \( q \) outside the shell at a distance \( r > R \) feels a force due to this electric field. The electric field \( E \) at a distance \( r \) from the center is given by \[ E = \frac{-Q}{4\pi\varepsilon_0 r^2}. \]The force \( F \) on the charge \( q \) is then \[ F = qE = \frac{-Qq}{4\pi\varepsilon_0 r^2}. \]This force is directed radially inward towards the center of the shell.
02

Understanding Force Inside the Shell

For \( r < R \), inside a uniformly charged spherical shell, by Gauss's law, the electric field is zero because any Gaussian surface inside the shell encloses no charge. Consequently, the force \( F \) experienced by the charge \( q \) is also zero since \[ F = q \times 0 = 0. \]

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in electrostatics. It represents the region around a charged object where other charges experience a force. The electric field, denoted by \( E \), is a vector quantity that points away from positive charges and towards negative charges.

Mathematically, the electric field arising from a charge \( Q \) is defined as the force \( F \) per unit positive test charge \( q \), or \( E = \frac{F}{q} \). In a spherical charge distribution, the electric field varies with distance from the charge.

For a point outside a charged sphere, the field behaves as if all the charge were concentrated at the center. For recomputing forces, consider the magnitude of the electric field using Gauss's Law, which depends inversely on the square of the distance from the center of the charge. This is critical when calculating the effects around a spherical distribution.
Spherical Charge Distribution
Spherical charge distribution involves spreading a total charge uniformly over the surface of a sphere. This setup presents ideal conditions for applying Gauss's Law due to its symmetry.

For a spherical shell with radius \( R \), if the total charge \(-Q\) is uniformly distributed over the surface, the electric field outside the shell (at \( r > R \)) is the same as if all the charge were concentrated at the center of the sphere. This causes the electric field to depend on the distance \( r \) from the center.

Inside the shell (at \( r < R \)), the electric field becomes zero. This intriguing result occurs because any Gaussian surface drawn inside the shell captures no net charge, effectively making the interior a "charge-free" zone. This leads to no force experienced by charges placed inside the shell.
Force on a Point Charge
The force on a point charge placed in an electric field is determined by integrating the effects of the field on the charge. When a positive point charge \( q \) is placed in a field created by a negative spherical charge distribution, it experiences a measurable force.

If this point charge is placed outside the spherical shell (\( r > R \)), the force \( F \) can be found using \( F = qE \), where \( E \) is the electric field due to the charged sphere. The resulting force is given by \( F = \frac{-Qq}{4\pi\varepsilon_0 r^2} \), directed radially towards the sphere's center as the fields attract oppositely charged particles.

Conversely, when the point charge is placed inside the hollow shell (\( r < R \)), it experiences zero force. This can be attributed to the absence of an electric field within the shell, as discussed under the concept of the spherical charge distribution.
Uniform Charge Distribution
A uniform charge distribution implies that charge is spread out evenly over a surface or volume, leading to consistent charge density. This simplification helps in calculations involving electric fields and forces.

In the case of a spherical insulating shell with a uniform distribution of charge \(-Q\), the symmetry allows for straightforward application of Gauss's Law. For points outside the sphere, the uniform distribution aids in treating the entire charge as if concentrated at the center.

This ensures the electric field diminishes with the square of the distance from the center when calculating resultant forces. Inside the shell, however, uniformity results in zero net electric field, creating a region where internal charges are unaffected by the sphere's charge. This principle illustrates why uniform charge distributions are essential in simplifying electric field computations.

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Most popular questions from this chapter

(a) At a distance of 0.200 \(\mathrm{cm}\) from the center of a charged conducting sphere with radius \(0.100 \mathrm{cm},\) the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 0.600 \(\mathrm{cm}\) from the center of the sphere? (b) At a distance of 0.200 \(\mathrm{cm}\) from the axis of a very long charged conducting cylinder with radius \(0.100 \mathrm{cm},\) the electric field is 480 \(\mathrm{N} / \mathrm{C}\) . What is the electric field 0.600 \(\mathrm{cm}\) from the axis of the cylinder? (c) At a distance of 0.200 \(\mathrm{cm}\) from a large uniform sheet of charge, the electric field is 480 \(\mathrm{N} / \mathrm{C} .\) What is the electric field 1.20 \(\mathrm{cm}\) from the sheet?

A solid metal sphere with radius 0.450 \(\mathrm{m}\) carries a net charge of 0.250 \(\mathrm{nC}\) . Find the magnitude of the electric field (a) at a point 0.100 \(\mathrm{m}\) outside the surface of the sphere and (b) at a point inside the sphere, 0.100 \(\mathrm{m}\) below the surface.

At time \(t=0\) a proton is a distance of 0.360 \(\mathrm{m}\) from a very large insulating sheet of charge and is moving parallel to the sheet with speed \(9.70 \times 10^{2} \mathrm{m} / \mathrm{s}\) . The sheet has uniform surface charge density \(2.34 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2} .\) What is the speed of the proton at \(t=5.00 \times 10^{-8} \mathrm{s} ?\)

A small conducting spherical shell with inner radius \(a\) and outer radius \(b\) is concentric with a larger conducting spherical shell with inner radius \(c\) and outer radius \(d\) (Fig. P22.47). The inner shell has total charge \(+2 q,\) and the outer shell has charge \(+4 q .\) (a) Calculate the electric field (magnitude and direction) in terms of \(q\) and the distance \(r\) from the common center of the two shells for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(b < r < c ;\) (iv) \(c < r < d\) ; (v) \(r > d .\) Show your results in a graph of the radial component of \(\vec{\boldsymbol{E}}\) as a function of \(r .\) (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (ii) inner surface of the large shell; (iv) outer surface of the large shell?

A region in space contains a total positive charge \(Q\) that is distributed spherically such the volume charge density \(\rho(r)\) is given by $$\begin{array}{ll}{\rho(r)=\alpha} & {\text { for } r \leq R / 2} \\\ {\rho(r)=2 \alpha(1-r / R)} & {\text { for } R / 2 \leq r \leq R} \\\ {\rho(r)=0} & {\text { for } r \geq R}\end{array}$$ Here \(\alpha\) is a positive constant having units of \(\mathrm{C} / \mathrm{m}^{3} .\) (a) Determine \(\alpha\) in terms of \(Q\) and \(R .\) (b) Using Gauss's law, derive an expression for the magnitude of \(\vec{\boldsymbol{E}}\) as a function of \(r .\) Do this separately for all three regions. Express your answers in terms of the total charge \(Q .\) Be sure to check that your results agree on the boundaries of the regions. (c) What fraction of the total charge is contained within the region \(r \leq R / 2 ?(\mathrm{d})\) If an electron with charge \(q^{\prime}=-e\) is oscillating back and forth about \(r=0\) (the center of the distribution) with an amplitude less than \(R / 2,\) show that the motion is simple harmonic. Hint: Review the discussion of simple harmonic motion in Section \(14.2 .\) If, and only if, the net force on the electron is proportional to its displacement from equilibrium, the motion is simple harmonic.) (e) What is the period of the motion in part (d)? If If the amplitude of the motion described in part (e) is greater than \(R / 2,\) is the motion still simple harmonic? Why or why not?

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