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Chapter 22: Problem 22

A point charge of -3.00 \(\mu\)C is located in the center of a spherical cavity of radius 6.50 cm that, in turn, is at the center of an insulating charged solid sphere. The charge density in the solid is \(\rho =\) 7.35 \(\times\) 10\(^{-4}\) C/m\(^3\). Calculate the electric field inside the solid at a distance of 9.50 cm from the center of the cavity.

Short Answer

Expert verified
The electric field inside the sphere at 9.50 cm is determined by calculating the net charge enclosed and applying Gauss's Law.

Step by step solution

01

Understanding the Problem

We need to find the electric field inside a solid sphere with a point charge at its center. The sphere has a uniform volume charge density and we need to consider both the charge of the point and the charge due to the sphere.
02

Calculate Enclosed Charge

Firstly, the charge inside the sphere up to a radius of 9.50 cm can be calculated using the charge density \( \rho \) and the volume of the sphere: \( Q = \rho \cdot \frac{4}{3}\pi (0.095)^3 \). Evaluate this to get the enclosed charge from the sphere.
03

Consider the Point Charge

Since the point charge at the center is -3.00 \( \mu \)C, which equals \(-3.00 \times 10^{-6} \) C, we need to take this into account as part of the enclosed charge. The total enclosed charge \( Q_{enc} \) is the charge due to the sphere plus the point charge.
04

Apply Gauss's Law

Using Gauss's Law, the electric field at any point inside the sphere is given by \( E = \frac{Q_{enc}}{4\pi\varepsilon_0 r^2} \), where \( r = 9.50 \) cm is the distance from the center, and \( \varepsilon_0 = 8.85 \times 10^{-12} \) C虏/(N路m虏) is the permittivity of free space. Substitute the values to find \( E \).
05

Solve for Electric Field

Calculate the total enclosed charge and substitute it into the Gauss's Law equation. Evaluate the expression to find the electric field at the given distance.

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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 electromagnetism, which describes how electric forces are exerted in the space surrounding electric charges. It's a vector field, meaning it has both magnitude and direction. The electric field
  • is created by electric charges
  • can exert force on other nearby charges
  • is strongest near the charge that creates it
The strength of the electric field (\( E \)) created by a point charge (\( Q \)) at a distance (\( r \)) can be calculated using the formula: \[E = \frac{Q}{4\pi\varepsilon_0 r^2}\] where \( \varepsilon_0 \) is the permittivity of free space. This helps in understanding how influence spreads from the charge outward.
Charge Density
Charge density (\( \rho \)) is the amount of electric charge per unit volume in a region of space. It helps us comprehend how charge is distributed within a material. In our problem, the charge density of the solid sphere is given as 7.35 \times 10^{-4} C/myard鲁. This uniform distribution means:
  • The charge is spread evenly throughout the sphere's volume.
  • Calculations using this charge density are simplified, as simple geometric formulas can be used to find the enclosed charge.
  • It affects how the electric field behaves inside insulators or conductors, impacting how charges influence one another in these materials.
Enclosed Charge
To find the electric field at a specified point, the concept of enclosed charge (\( Q_{enc} \)) is essential. It's the total charge enclosed within a given volume. For this exercise:
  • Calculate the charge enclosed by the sphere at a radius of 9.50 cm using the equation: \( Q = \rho \cdot \frac{4}{3}\pi(0.095)^3 \)
  • Include the contribution from the central point charge of -3.00 \( \mu \)C.This concept is crucial because Gauss's Law requires knowing all the charges affecting the region of interest to determine the electric field effectively.
Point Charge
A point charge is an idealization where a charge is assumed to be concentrated at a single point in space. In practice, it provides ease in calculations and theoretical models. In this problem:
  • The given point charge is -3.00 \( \mu \)C, centrally located within the sphere.
  • Its negative sign indicates that it will exert an attractive force on positive charges.
  • This singular charge significantly influences the local electric field inside the surrounding material, and its effects are considered separately when calculating electric fields using Gauss's Law.
Permittivity of Free Space
The permittivity of free space (\( \varepsilon_0 \)) is a constant that characterizes the ability of the vacuum to permit electric field lines. Its value is approximately 8.85 \times10^{-12} C虏/(N路m虏). This constant appears frequently in equations involving electric fields and forces.
  • It defines the strength of the electric field generated per unit charge in a vacuum.
  • Essential for calculations involving electric fields as it bridges the conceptual understanding of fields with practical calculations.
  • Frequently used in Gauss's Law to relate enclosed charge to the electric field.

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Most popular questions from this chapter

A 6.20-\(\mu\)C point charge is at the center of a cube with sides of length 0.500 m. (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 m long? Explain.

It was shown in Example 21.10 (Section 21.5) that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude \(E = \lambda/2\pi\varepsilon_0r\). Consider an imaginary cylinder with radius \(r =\) 0.250 m and length \(l =\) 0.400 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is \(\lambda =\) 3.00 \(\mu\)C/m. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to \(r =\) 0.500 m? (c) What is the flux through the cylinder if its length is increased to \(l =\) 0.800 m?

(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cm in diameter to produce an electric field of magnitude 1390 N/C just outside the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

A region in space contains a total positive charge \(Q\) that is distributed spherically such that the volume charge density \(\rho(r)\) is given by $$\rho(r) = 3ar/2R \space \space \space \space \space \mathrm{for} \space r \leq R/2$$ $$\rho(r) = \alpha[1-(r/R)^2] \space \space \space \space \mathrm{for} \space R/2 \leq r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \mathrm{for} \space r \geq R$$ Here \(\alpha\) is a positive constant having units of C/m\(^3\). (a) Determine \(\alpha\) in terms of \(Q\) and \(R\). (b) Using Gauss's law, derive an expression for the magnitude of the electric field as a function of \(r\). Do this separately for all three regions. Express your answers in terms of \(Q\). (c) What fraction of the total charge is contained within the region \(R/2 \leq r \leq R\)? (d) What is the magnitude of \(\overrightarrow{E}\) at \(r = R/2\)? (e) If an electron with charge \(q' = -e\) is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why?

A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude 90.0 N/C that is directed at 20\(^\circ\) from the plane of the sheet (\(\textbf{Fig. E22.2}\)). Find the magnitude of the electric flux through the sheet.
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