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Learning objectives

By the end of this section, you will be able to:

  • List the three properties of a conductor in electrostatic equilibrium.
  • Explain the effect of an electric field on free charges in a conductor.
  • Explain why no electric field may exist inside a conductor.
  • Describe the electric field surrounding Earth.
  • Explain what happens to an electric field applied to an irregular conductor.
  • Describe how a lightning rod works.
  • Explain how a metal car may protect passengers inside from the dangerous electric fields caused by a downed line touching the car.

The information presented in this section supports the following AP learning objectives:

  • 2.C.3.1 The student is able to explain the inverse square dependence of the electric field surrounding a spherically symmetric electrically charged object.
  • 2.C.5.1 The student is able to create representations of the magnitude and direction of the electric field at various distances (small compared to plate size) from two electrically charged plates of equal magnitude and opposite signs and is able to recognize that the assumption of uniform field is not appropriate near edges of plates.

Conductors contain free charges that move easily. When excess charge is placed on a conductor or the conductor is put into a static electric field, charges in the conductor quickly respond to reach a steady state called electrostatic equilibrium    .

[link] shows the effect of an electric field on free charges in a conductor. The free charges move until the field is perpendicular to the conductor's surface. There can be no component of the field parallel to the surface in electrostatic equilibrium, since, if there were, it would produce further movement of charge. A positive free charge is shown, but free charges can be either positive or negative and are, in fact, negative in metals. The motion of a positive charge is equivalent to the motion of a negative charge in the opposite direction.

In part a, an electric field E exists at some angle with the horizontal applied on a conductor. One component of this field E parallel is along x axis represented by a vector arrow and other E perpendicular, is along y axis represented by a vector arrow. Charge inside the conductor moves along x axis so the force acting on it is F parallel, which is equal to q multiplied by E parallel. In part b, a charge is shown inside the conductor and electric field is represented by a vector arrow pointing upward starting from the surface of the conductor.
When an electric field E size 12{E} {} is applied to a conductor, free charges inside the conductor move until the field is perpendicular to the surface. (a) The electric field is a vector quantity, with both parallel and perpendicular components. The parallel component ( E size 12{E rSub { size 8{ \rdline } } } {} ) exerts a force ( F size 12{F rSub { size 8{ \rdline } } } {} ) on the free charge q size 12{q} {} , which moves the charge until F = 0 size 12{F rSub { size 8{ \rdline } } =0} {} . (b) The resulting field is perpendicular to the surface. The free charge has been brought to the conductor's surface, leaving electrostatic forces in equilibrium.

A conductor placed in an electric field will be polarized    . [link] shows the result of placing a neutral conductor in an originally uniform electric field. The field becomes stronger near the conductor but entirely disappears inside it.

A spherical conductor is placed in the external electric field. The field lines are shown running from left to right. The field lines enter and leave the conductor at right angles. Negative charges accumulate on the left surface of the conductor and positive charges accumulate on the right surface of the conductor.
This illustration shows a spherical conductor in static equilibrium with an originally uniform electric field. Free charges move within the conductor, polarizing it, until the electric field lines are perpendicular to the surface. The field lines end on excess negative charge on one section of the surface and begin again on excess positive charge on the opposite side. No electric field exists inside the conductor, since free charges in the conductor would continue moving in response to any field until it was neutralized.

Questions & Answers

sound waves can be modeled as a change in pressure ,why is the change on in pressure used and not the actual pressure
Dotto Reply
what is the best
Kelly Reply
Water,air,fire
Maung
I am a university student of Myanmar.I am first year,first semester.I want to learn about physics.
Maung
two charges qA and qB are separated by a distance x. if we double the distance between the charges and triple the magnitude of the charge A, what happens to the magnitude of the force that charge A exerts on charge B. what happens to the magnitude of the force that charge B exerts on charge A
tanla Reply
how to get mcq and essay?
Owen Reply
what is force
Ibrahim Reply
force is a pull or push action on an object or a body.
joseph
what is a significant figure? and give example
Frederick
numerical chapter number 3
Sajid Reply
joined
Ibrahim
a reflected ray on a mirror makes an angle of 20degree with the incident ray when the mirror is rotated 15degree what angle will the incident ray now make with the reflected ray
Akinyemi Reply
what is simple harmonic motion
Solomon Reply
how vapour pressure of a liquid lost through convection
Yomzi Reply
Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena.
Aliraza Reply
Plz answer the question ☝️☝️
Aliraza
what's the basic si unit of acceleration
ELLOIN Reply
Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.
Fabian Reply
Insulators (nonmetals) have a higher BE than metals, and it is more difficult for photons to eject electrons from insulators. Discuss how this relates to the free charges in metals that make them good conductors.
Muhammad Reply
Is the photoelectric effect a direct consequence of the wave character of EM radiation or of the particle character of EM radiation? Explain briefly.
Muhammad
Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
Henny Reply
how solve this problem?
Foday
P(pressure)=density ×depth×acceleration due to gravity Force =P×Area(28.0x8.5)
Fomukom
for the answer to complete, the units need specified why
muqaddas Reply
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
Practice Key Terms 6

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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