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Chapter 22: Problem 16

22.16. 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 \((\mathrm{b})\) at a point inside the sphere, 0.100 \(\mathrm{m}\) below the surface.

Short Answer

Expert verified
Outside: 7.42 N/C, Inside: 0 N/C.

Step by step solution

01

Understand the Problem

We are given a solid metal sphere with a radius of 0.450 m carrying a net charge of 0.250 nC. We need to find the electric field at certain points relative to this sphere.
02

Determine the Field Outside the Sphere

The electric field at a point outside a charged metal sphere can be treated as if all the sphere's charge is concentrated at its center. Use the formula for the electric field due to a point charge: \[ E = \frac{kQ}{r^2} \]where \( k = 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant), \( Q = 0.250 \times 10^{-9} \, \text{C} \), and \( r = 0.450 + 0.100 = 0.550 \, \text{m} \).
03

Calculate the Electric Field Outside

Substitute the values into the formula:\[ E = \frac{8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \times 0.250 \times 10^{-9} \, \text{C}}{0.550^2 \, \text{m}^2} \]Calculate to find the magnitude of the electric field.
04

Result for the Electric Field Outside

After calculation, the electric field at 0.100 m outside the surface is approximately 7.42 N/C.
05

Determine the Field Inside the Sphere

Inside a conducting sphere, the electric field is zero. The charges on the surface cause any point within to have no net electric field due to the symmetrical distribution of charge.
06

Result for the Electric Field Inside

The electric field at 0.100 m inside the sphere is 0 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 of electromagnetism that describes the force between two charged objects. It states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This can be expressed with the formula: \[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\]where:
  • \( F \) is the force between the charges,
  • \( k \) is Coulomb's constant \( (8.988 \times 10^9 \, ext{N m}^2/ ext{C}^2) \),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
  • \( r \) is the distance between the charges.
The force is attractive if the charges have opposite signs and repulsive if they have the same sign. This principle is crucial when determining electric forces and fields for both point charges and continuous charge distributions.
Conducting Sphere
A conducting sphere is an important subject in electrostatics. It is made of a material that allows electric charges to move freely across its surface. When a charge is placed on a conducting sphere, the charge uniformly distributes itself over the surface due to repulsion among like charges. The result is:
  • The electric charge is only present on the surface of the sphere.
  • Within the sphere, the electric field is zero because the charges distribute themselves symmetrically, canceling out any internal fields.
  • The sphere can be treated as a point charge when calculating the electric field outside it, with the entire charge effectively acting as if it is located at the center.
These properties make conducting spheres ideal for understanding fundamental electrostatic principles.
Point Charge
A point charge is an idealized model in physics used to simplify and understand electric fields and forces. It is a hypothetical charge with an infinitely small size, so its shape doesn't affect calculations. Point charges are useful when:
  • Analyzing electric fields generated by simple charge distributions.
  • Using Coulomb's law to calculate forces and fields around charged objects, treating the charge as located at a single point.
  • Imagining the charge as concentrated in a very small region, particularly useful in symmetry arguments and under conditions where the exact location doesn't change the outcome, as seen in the electric field calculations for spherical distributions.
Point charges simplify complex electrostatic problems, especially when dealing with spherical objects, as all external electric effects can be computed as if the charge were collected at one central point.
Electric Field Inside a Conductor
Understanding the electric field inside a conductor is vital for grasping electrostatic equilibrium concepts. In the electrostatic state (where charges are stationary), the electric field inside a conductor must be zero. Any non-zero electric field would cause charges within the conductor to move, disrupting this equilibrium. Considerations include:
  • Electrostatic shielding: A conductor can shield its interior from external electric fields.
  • Charge distribution occurs on the surface, ensuring no internal electric field.
  • Because of this, if you were to look for an electric field within the body of a conducting sphere, you'd find none.
This principle is applied in practice, such as in Faraday cages, which protect sensitive equipment from external electric fields.

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Most popular questions from this chapter

22.63. Positive charge \(Q\) is distributed uniformly over each of two spherical volumes with radius \(R\) . One sphere of charge is centered at the origin and the other at \(x=2 R\) (Fig. 2244 ). Find the magnitude and direction of the net electric field due to these two distributions of charge at the following points on the \(x\) -axis: (a) \(x=0 ;\) (b) \(x=R / 2 ;(c) x=R ;\) (d) \(x=3 R\) .

22.37. The Coasial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c\) . The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \(\lambda\) . Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and \((b)\) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r=0\) to \(r=2 c\) (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

22.54. A Uniformiy Charged Stab. A slab of insulating material has thickness 2\(d\) and is oriented so that its faces are parallel to the \(y z\) -plane and given by the planes \(x=d\) and \(x=-d\) . The \(y\) - and \(z\) -dimensions of the slab are very large compared to \(d\) and may be treated as essentially infinite. The slab has a uniform positive charge density \(\rho\) . (a) Explain why the electric field due to the slab is zero at the center of the slab \((x=0)\) . (b) Using Gauss's law, find the electric field due to the slab (magnitude and direction) at all points in space.

22.45. Concentric Spherical Shells. 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. 22.39\()\) . 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) \(rd\) . Show your results in a graph of the radial component of \(\vec{E}\) 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; (iii) inner sur-=face of the large shell; (iv) outer surface of the large shell?

22.25. The electric field at a distance of 0.145 \(\mathrm{m}\) from the surface of a solid insulating sphere with radius 0.355 \(\mathrm{m}\) is 1750 \(\mathrm{N} / \mathrm{C}\) . (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? ( 6 ) Calculate the electric field inside the sphere at a distance of 0.200 \(\mathrm{m}\) from the center.
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