Conductors in electrostatics In a conductor electrons are free to mo ve. If they are acted on by a force, they will acce lerate in the direction of the force. If a conductor is placed into an external electric field, a force F = -eEacts on each free electron. Electrons acce lerate and g ain velocity in a direction opposite to the field. Soon electrons will pile up on the surface on one side of the conductor, while the surface on the other side will be depleted ofelectrons and have a ne t positive charge. These separated neg ative and positive charges o n opposing sides of the conductor produce their own electric field, which opposes the external field inside the conductorand modifies the field outside. When enough electrons have piled up on one side and enough positive charge has been left on the otherside, then the field produced by these charges exactly cancels the external field inside the conductor, and electrons inside the cond uctor no longer experience a force. This is the case in the picture sho wn above. The inside of the conducting sphere is field-free, while the previously constant external field outside has been modified. In static equilibrium the inside of a conductor is field free. If it were not, electrons would move and distribute themselves, so as to cancel o ut the field. The inside of a conductor can n ot contain any n et charge. Such charges would produce a field inside the conductor, and electrons would move and cancel out t he field and neutralize the charge. Any exces s charge on a conductor must therefore reside on the surface. The field just outside the conductor at the s urface must be perpendicular to the s urface. If it were not, electrons would redistribute themselves to canc el out the field. The strength of the electric field on the surfac e of a conductorcan be found by applying Gauss' law. The electric flux through the surface shown in the figure above is Φ = Q inside /ε 0 = σ A/ε 0 , where σis the surface charge density a nd A is the area of the conductor's surfac e inside the Gaussian surfac e shown. The flux through the sides of the Gaussian surface is zero, since Eis perpendicular to the surface of the conductor. The flux through the bottom of the Gaus sian surface is zero since the e lectric field inside a conductor is zero. The total flux through the Gaussian surface therefore equals the flux throug h the top, and
n a conductor electrons are free to move. If they are acted on by a force, they will accelerate in the directionf the force. If a conductor is placed into an external electric field, a force F = -eE acts on each freelectron. Electrons accelerate and gain velocity in a direction opposite to the field. Soon electrons will pile
p on the surface on one side of the conductor, while the surface on the other side will be depleted of lectrons and have a net positive charge. These separated negative and positive charges on opposingdes of the conductor produce their own electric field, which opposes the external field inside the conductor nd modifies the field outside.
When enough electrons have piled up on one side and enough positive charge has been left on the other de, then the field produced by these charges exactly cancels the external field inside the conductor, and
lectrons inside the conductor no longer experience a force. This is the case in the picture shown above.he inside of the conducting sphere is field-free, while the previously constant external field outside haseen modified.
n static equilibrium the inside of a conductor is field free. If it were not, electrons would move and distributehemselves, so as to cancel out the field. The inside of a conductor can not contain any net charge. Suchharges would produce a field inside the conductor, and electrons would move and cancel out the field andeutralize the charge. Any excess charge on a conductor must therefore reside on the surface. The field
ust outside the conductor at the surface must be perpendicular to the surface. If it were not, electrons wouldedistribute themselves to cancel out the field. The strength of the electric field on the surface of a conductor an be found by applying Gauss' law.
he electric flux through the surface shown in the figure above is Φ = Qinside/ε0 = σ A/ε0, where σ is the
urface charge density and A is the area of the conductor's surface inside the Gaussian surface shown. The
ux through the sides of the Gaussian surface is zero, since E is perpendicular to the surface of theonductor. The flux through the bottom of the Gaussian surface is zero since the electric field inside aonductor is zero. The total flux through the Gaussian surface therefore equals the flux through the top, and
we have Φ = EA. We therefore have for the strength of the electric field near the surface of a conductor
E = σ/ε0.
he surface of any conductor is an equipotential surface. The field is everywhere perpendicular to theurface. No work is being done moving a charge along on the surface. The surface of a spherical conductor
with radius R, carrying a charge Q is at a potential V = keQ/R. If we have two spherical conductors with radii
R1 and R2, respectively, at the same potential V, they carry charges Q1 = R1V/k and Q2 = R2V/k,
espectively. The electric fields near their surfaces are E1 = keQ1/R12
= V/R1 and E2 = keQ2/R22
= V/R2
espectively. The smaller the radius, the larger is the electric field. In general, near the surface of aonductor , the field is largest in places with the smallest local radius of curvature.
ery strong fields are found near sharp conducting tips. Air molecules will be stripped of electrons if the field
ecomes too large (~3×106 V/m). The free electrons accelerate and collide with other molecules to makemore ions and electrons. A plasma forms between the conductor and the ground and the conductor
ischarges. This is called a corona discharge.
device that makes use of the strong field near a tip is the field ion microscope. The field ion microscopeas a sharp tip with a local radius of curvature of ~10 - 100 nm. This tip faces a phosphor screen. Under acuum, a potential difference is established between the tip and the screen. The tip is held at the more
ositive potential. A small amount of inert gas is admitted, and gas atoms near the tip are ionized. Electronsre ripped off these atoms. The positively charged ions are accelerated by the intense electric field along atraight line toward the phosphor screen, where they are detected by converting their kinetic energy intoght. Each point on the tip maps into a different point on the screen, so that a magnified, image of the tipan be viewed. Since the tip is only 10 - 100nm in radius, one can achieve atomic resolution. A typical field
on microscope image of a 'single crystal' tungsten tip is shown below:
he bright spots correspond to positions on the tip where the electric field is particularly high, i.e. where theocal radius of curvature is particularly small. This happens near atoms, so the microscope images the
osition of atoms in the tip.
ink:
The ORNL Atom Probe
ummary:
n electrostatic equilibrium a conductor has the following properties.
Any excess charge resides on the surface of the conductor.
The electric field is zero within the solid part of the conductor.
The electric field at the surface of the conductor is perpendicular to the surface.
Charge accumulates, and the field is strongest, on pointy parts of the conductor.
he entire conductor is at the same potential. There is no field inside the conductor. A cavity inside a
onductor, completely surrounded by conducting material, also is free of electric fields, if it does not containny net charge itself. A conductor shields its interior from any outs ide electric fields. Even if there areoles in the surface, the electric field does not penetrate very far. A rule of thumb is that the electric field
alls to zero over a distance approximately equal to the diameter of the hole.
n the diagram above, the field only penetrates a small distance through the holes into the box withonducting walls.
Why are you safest inside your car during a thunderstorm?
roblem:
A square plate of copper with 50cm sides has no net charge and is placed in a region of uniformelectric field of 80 kN/C directed perpendicular to the plate. Find(a) the charge density on each face of the plate and(b) the total charge on each phase.
Solution:(a) The surface charges on the plate must produce a field, which exactly cancels the
external field inside the conductor. The magnitude of the field produce by a surface chargeis E = σ/2ε0. We require that the field produced by the charges on the upper and the field
produced by the charges on the lower surface together cancel the external field. Thecharge density on one of the surfaces is positive. On the other surface the charge densityis negative it has the same magnitude, since the net charge is zero. Inside the conductor the fields produced by the charges on the two surfaces have the same direction and we
add their magnitudes to find the magnitude of the total field. We need Eext = 2E = σ/ε0.
We therefore need σ = ε0×80kN/C = 7×10-7
C/m2 for the magnitude of the charge density.
One plate has positive, and on plate has negative charge density.
(b) The magnitude of total charge on each plate is Q =σ A = (7
×10
-7
C/m
2
)×0.25m
2
=1.77×10
-7C. One plate has positive and one plate has negative charge.
apacitors
capacitor is a device for storing charge. No single electronic component plays a more important role todayhan the capacitor. This device is used to store information in computer memories, to regulate voltages inower supplies, to establish electrical fields, to store electrical energy, to detect and producelectromagnetic waves, and to measure time.
ny two conductors separated by an insulating medium form a capacitor. A parallel plate capacitor consists
f two plates separated by a thin insulating material known as a dielectric. One plate of the capacitor holdspositive charge Q, while the other holds a negative charge -Q. The charge Q on the plates is proportional
o the potential difference V across the two plates. The capacitance C is the proportional constant,
depends on the capacitor's geometry and on the type of dielectric material used. The capacitance of aarallel plate capacitor with two plates of area A separated by a distance d and no dielectric materialetween the plates is
C = ε0 A/d.
The electric field is E = σ/ε0. The voltage is V = Ed = σd/ε0. The charge is Q = σ A. Therefore Q/V = σ Aε0/σd
Aε0/d.) The SI unit of capacitance is Coulomb/Volt = Farad (F). Typical capacitors have capacitances in
he picoFarad to microFarad range.
he capacitance tells us how much charge the device stores for a given voltage. A dielectric between theonductors increases the capacitance of a capacitor. The molecules of the dielectric material are polarized
n the field between the two conductors. The entire negative and positive charge of the dielectric isisplaced by a small amount with respect to each other. This results in an effective positive surface chargen one side of the dielectric and a negative surface charge on the other side of the dielectric. Theseffective surface charges on the dielectric produce an electric field, which opposes the field produced by theurface charges on the conductors, and thus reduces the voltage between the conductors. To keep theoltage up, more charge must be put onto the conductors. The capacitor thus stores more charge for a given
oltage. The dielectric constant κ is the ratio of the voltage V0 between the conductors without the dielectric
o the voltage V with the dielectric, κ = V0/V, for a given amount of charge Q on the conductors.
n the diagram above, the same amount of charge Q on the conductors results in a smaller field between the
lates of the capacitor with the dielectric. The higher the dielectric constant κ, the more charge a capacitor an store for a given voltage. For a parallel-plate capacitor with a dielectric between the plates, the
apacitance is
C = Q/V = κQ/V0 = κε0 A/d.
he static dielectric constant of any material is always greater than 1.
he energy stored in a capacitor is equal to the work done in separating the charges on the conductors. Themore charge is already stored on the plates, the more work must be done to separate additional charges,
ecause of the strong repulsion between like charges. At a given voltage, it takes an infinitesimal amount of work dW = VdQ to separate an additional infinitesimal amount of charge dQ. (The voltage V is the amount
f work per unit charge.) We can write dW = (Q/C)dQ, since V = Q/C. To find the total work done in
harging the capacitor we integrate,
.
he energy stored in a capacitor therefore is
U = (1/2)(Q2/C).
sing Q = CV we can also write
U = (1/2)CV2.
ink:
Variable Capacitor Exercise
roblem:
Each memory cell in a computer contains a capacitor to store charge. Charge being stored or not
being stored corresponds to the binary digits 1 and 0. To pack the cells more densely, trenchcapacitors are often used in which the plates of a capacitor are mounted vertically along the walls of a
trench etched into a silicon chip. If we have a capacitance of 50 femtoFarad = 50×10-15
F and each
plate has an area of 20×10-12m2 (micron-sized trenches), what is the plate separation?Solution:
If there is no dielectric material, then
C = ε0 A/d, d = ε0 A/C = (8.85×10-12
×20×10-12
/(50×10-15
))m = 3.54×10-9
m.
Typical atomic dimensions are on the order of 0.1nm, so the trench is on the order of 30 atomswide.
or any insulator, there is a maximum electric field that can be maintained without ionizing the molecules.or a capacitor this means that there is a maximum allowable voltage that that can be placed across theonductors. This maximum voltage depends the dielectric in the capacitor. The corresponding maximumeld is called the dielectric s trength of the material. For stronger fields, the capacitor 'breaks down' (similar
o a corona discharge) and is normally destroyed. Most capacitors used in electrical circuits carry both aapacitance and a voltage rating. This breakdown voltage Vb is related to the dielectric strength Eb. For a
arallel plate capacitor we have Vb = Ebd.
Material Dielectric Strength (V/m)
Air 3×106
Bakelite 24×106
Neoprene rubber 12×106
Nylon 14×106
Paper 16×106
Polystyrene 24×106
Pyrex glass 14×106
Quartz 8×106
Silicone oil 15×106
Strontium titanate 8×106
Teflon 60×106
apacitors in parallel
onsider two capacitors in parallel as shown below.
When the battery is connected, electrons will flow until the potential of point A is the same as the potential of he positive terminal of the battery and the potential of point B is equal to that of the negative terminal of theattery. Thus, the potential difference between the plates of both capacitors is V A - VB = Vbat. We have C1
Q1/Vbat and C2 = Q2/Vbat, where Q1 is the charge on capacitor C1, and Q2 is the charge on capacitor C2.
et C be the equivalent capacitance of the two capacitors in parallel, i.e. C = Q/Vbat, where Q = Q1 + Q2.hen C = (Q1 + Q2)/Vbat = C1 + C2. For capacitors in parallel, the capacitances add. For more than two
et Q represent the total charge on the top plate of C1, which then induces a charge -Q on its bottom plate.
he charge on the bottom plate of C2 will be -Q, which in turn induces a charge +Q on its top plate as
hown.
et V1 and V2 represent the potential differences between plates of capacitors C1 and C2, respectively.
hen V1 + V2 = Vbat, or (Q/C1) + (Q/C2) = Q/C, or (1/C1) + (1/C2) = 1/C. For more than two capacitors in
eries we have
where C is equivalent capacitance of the two capacitors.
lectrostatic air cleaners
lectrostatic air cleaners remove dust, soot, and ash particles from normal air. Each dust, soot, or asharticle has mass. The air exerts two types of forces on the particles, the buoyant force and the viscousrag force. Since the particles are denser than air, the buoyant force alone cannot support the particles.he viscous drag force keeps the particles from descending quickly. For small particles the terminal velocityan be lower than 1mm/s. The drag force opposes relative motion between the particles and the air, and
moving air tends to carry the particles along with it. The slightest upward breeze can keep the dust, soot,
nd ash particle aloft.
lectrostatic air cleaners use electrostatic forces to pull these particles from the air. A typical air cleaner ives each particle a negative charge and then collects it on a positively charged surface.
ow does a dust grain become negatively charged?
he air cleaner uses a corona discharge to give the dust grain a negative charge. A power supply does
work maintaining a potential difference of approximately 10000V between the corona wires and theollecting surfaces. The negatively charged dust flows with the air through the air cleaner. When it passes aositively charged surface, it experiences an electrostatic force strong enough to overwhelm the viscousrag. The dust particles quickly leave the air stream and collect on the charged surface. The air continuesn without the dust. The air cleaner precipitates clumps of dust on its collecting plates, and therefore is
alled an electrostatic precipitator . It can accumulate large amounts of dust on its plates without blockinghe airflow and it is easy to clean. When several centimeters of dust have accumulated on the collectingurfaces, it is removed by rapping the plates with a stick. The sudden blow causes the plates to accelerateapidly and they leave the dust behind. It falls in clumps to the bottom of the precipitator, where it’s collectedor recycling or disposal.
on Generators
ousehold ion generators are also effective at removing dust and smoke from room air. These machinesesemble electrostatic precipitators, but they have no internal collecting plates. They use a coronaischarges to charge passing molecules and dust grains and then let those charged particles drift into theoom. When the charged particles come near a surface, they polarize the surface and are attracted to it.lthough this method is cheap and effective, it slowly dirties the walls and the furniture.
ink:
Electrostatic Precipitation
hotocopiers
t the heart of the photocopier is a thin layer of photoconductor . A photoconductor is a solid material
hrough which electrons can move only when it is exposed to light. In the dark, it is an insulator, in the
ght, it’s a conductor. This property allows light to determine the pattern of static electricity on a copyingrum and hence the placement of toner on a piece of paper.
he light sensitive component in a photocopier is a metal drum that is covered with a thin layer of hotoconductor. This metal drum is grounded. The copier coats the photoconductor with electrons, whichemain in place as long as the photoconductor is in the dark. But wherever light strikes the photoconductor,becomes conductive and allows the electrons to escape through the metal and flow into the ground. Only
he portions of the photoconductor which are not illuminated retain their static electric charge and eventuallyttract black toner particles. In that manner, the darkened parts of the photoconductor produce the darkarts of the final copy.
he copier starts by applying a uniform negative charge to the surface of the photoconductor. This charge ispplied by a corotron, a fine wire centered in a half-cylinder of metal. A power supply pumps electrons onto
he fine wire until they are emitted into the air as a corona discharge. When these electrons approach the
hotoconductor they polarize it and stick to it. The photoconductor becomes uniformly charged, with about
0-7C of negative charge per cm2 of surface. After the charging, the copier exposes the photoconductor toght from the original document. It uses a lens to cast an image of the original onto the photoconductor’surface. Light only hits the photoconductor in certain places, which correspond to the white parts of theriginal document. When the exposure is over, the photoconductor carries a charge image of the originalocument.
o develop this charge image into a visible one, the photocopier exposes the photoconductor to chargedoner particles. The toner is a fine insulating plastic powder. A spinning brush with extraordinarily softristles wipes toner particles out of their storage tray onto the photoconductor. During this transfer, the toner
articles become positively charged so that they stick to the negatively charged portions of thehotoconductor. The photoconductor now carries a black image of the original document. But to create aopy, this black image must be transferred to paper. To begin this transfer, the copier illuminates thehotoconductor with a charge erase lamp so that the photoconductor’s negative charge escapes into the
metal. The toner remains in place but it is only very weakly attached. The copier than transfers the toner tonearby sheet of paper by applying negative charge to the paper’s back. The positively charged toner isttracted to the negatively charged paper and the two leave the photoconductor together. The copier theneats and presses the copy, permanently fusing the toner into the paper. Sometimes, when a copier jams,
ou may remove a sheet before it has been fused. The image looks completely normal but wipes off whenou touch it because it’s held in place only by electrostatic forces. Once the image has been transferred tohe paper the drum is cleaned. The photoconductor is then ready to be used again.
laser printer is also a photocopier device, but it uses a laser beam to write a charge image directly onto itshotoconductor drum. Wherever laser light hits the drum, charge flows through the photoconductor. Aomputer in the printer turns the laser on and off as it systematically constructs the charge image, one dot attime. The photoconductor and the toner supply are contained in a single disposable cartridge.
Other links concerning electrostatics:
How does a photocopier work
How the Van de Graaff generator worksLightning
Lightning Primer
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