Transcript Electrostatics

Electrostatics
• Electostatics: the study of charges at rest.
• Charges give rise to electric force (electromagnetic
force, to be precise)
• Like charges repel, opposites attract.
• 2 types of charge: + and –
• In the majority of cases, only the negative charges are
mobile.
• An object that contains the same amnt of + and –
charges in close proximity attracts and repels an
external charge equally and thereby cancels it own
ability to exert a net force. It is neutral.
• An e- is negatively charged.
• There is no way to discharge an electron and make it
neutral. If the electron is a fundamental particle,
charge is an inseparable aspect of the thing itself.
• We have experimentally determined the charge of an
electron (qe). Electric charge comes in whole number
multiples of that basic amnt.
• Whether + or -, charge is quantized. It appears in
certain specific amnts.
• 0, ± qe, ± 2qe …
• A neutral object is electrified by either losing or
gaining charge; losing or gaining electrons.
Conservation of Charge
• The total charge (difference between the amnt of +
and – charge) within an isolated system is always
constant.
Charging by friction
• Ex: shuffling across the carpet in socks.
• Outer electrons are the least strongly bound and most
easily shed.
• Different materials have different affinities for
electrons.
• When in contact, one material may give up some of
its electrons to the other.
• Depends on the materials.
Triboelectric sequence
• “tribo”  friction
• Materials towards the top of the list tend to lose
electrons easily, those near the bottom tend to gain
them effectively.
• http://www.siliconfareast.com/tribo_series.htm
Transfer of + charge
• A positively charged object when placed near
something neutral will draw e- away from the neutral
object leaving both positively charged.
• ONLY e- are transferred. But the system behaves as
though positive charge flowed from the charged
object to the neutral.
Insulators and Conductors
• A conductor allows charge introduced anywhere
within it to flow freely and redistribute. (ex: metals)
• Good conductors hold onto their outer e- weakly.
• There is no perfect insulator, all allow some
redistribution of charge.
• Air is a good insulator (esp dry air).
• If enough negative charge builds up on an object,
electrons, under the force of mutual repulsion, may
be emitted into the surrounding air.
• This increases the amnt of free e- and ions in the air,
temporarily causing the air to become an electrical
conductor.
• The temperature increase causes some of the atoms to
emit light and sound (a spark).
Conduction
• A cluster of identical charges introduced on a
conductor experiences a mutual repulsion that sends
them scurrying until they can separate no farther and
are as far away from each other as possible.
• No matter what the shape of the conductor, excess
charge always resides on its outer surface.
• Another manifestation of the flow of e- in a conductor
is the manner in which charge is transferred.
• A negative conductor that touches an uncharged
metal body. e- are propelled onto the neutral body by
their mutual repulsion.
• Flow like a liquid until charge is equalized on both
objects. Equilibrium.
• Any good conductor in good contact with the Earth
(like water pipes in a building) will carry off all the
charge one could want to discard, depositing I into
the moist ground.
• To ground
Induction
• Charge is induced when no physical contact between
the objects.
Electric Force
• Mid 18th C.  question of the nature of the force
acting between charges was a major scientific issue.
• Speculation was very Newtonian, relying on
discoveries regarding gravity.
• Bernoulli, with his crude experiments, also confirmed
that electrical attraction/repulsion seemed to follow
an inverse square law.
• Ben Franklin was able to observe that there is no
charge on the inside wall of a hollow electrified
conductor. All charge was on the outside.
• All this led up to Charles Augustin de Coulomb’s
definitive study on electrical force (1785).
Coulomb’s Law
FE = k q1q2
2
r
• The force acts between two charges is directly
proportional to the product of the charges and
inversely proportional to the square of the
distance separating them.
• + force means repulsive, - means attractive
• Units for charge: coulomb (C)
• k =8.98755179 x 109 Nm2/C2
• The charge of an e- is -1.06217733 x 10-19 C
Example 1
A Hydrogen atom consists of an electron of mass
9.1094 x 10-31 kg, moving about a proton of mass
1.6276 x 10-27 kg at an avg distance of 0.53 x 10-10 m.
Find the electric and gravitational forces acting
between the two particles.
• Experimentally confirmed that when several charges
are present, each exerts a force given by Coulomb’s
Law on every other charge.
• The interaction btwn any two charges is
independent of the presence of all other charges.
• The net force on any one charge is the sum of all the
forces exerted on it due to each of the other charges
interacting with it independently.
Example 2
Drawing on board shows three tiny uniformly charged
spheres. Determine the net force acting on the middle
sphere due to the other two.
Example 3
Drawing on the board depicts three small charged
spheres at the vertices of a 3-4-5 right triangle.
Calculate the force (mag and dir) exerted on q3 by the
other two charges.
Electric Field
• A field of force exists in a region of space when an
appropriate object placed at any point therein
experiences a force.
• Michael Faraday was the first to introduce a visual
representation of the electric force field.
• Uses lines of force.
• The force experienced by a + test charge at any point
in space is in the direction tangent to the line offorce
at that point.
• Flow out of + and into –
• The more lines drawn in a region (i.e., the denser the
concentration of the lines), the stronger the field they
represent.
• The farther apart the lines, the weaker the field.
• We define the electric field (E with an arrow on top)
at a point in space to be the electric force experienced
by a + test charge at that point divided by that charge.
E = F/qo
• Force on a point charge could be found with same
eqn.
F = Eq
Example 4
• At a particular moment the electric field at a point 30
cm above an electric blanket is 250 N/C, straight up.
Compute the force that acts on an e- at that location,
at that moment.
• Electric field of a point charge
E = k (q/r2)
Allows us to calculate the E-field anywhere in space
due to a known distribution of point-charges (e-, p+,
whatever).
Each charge contributes a field to the overall E-field.
Example 5
• Drawing on the board shows two point-charges (or
tiny charged spheres) each of 10nC separated in air
by a distance of 8.0 m. Compute the electric field at
points A, B, and C.
• English electrical engineer, Oliver Heaviside, pointed
out around 1892 that most of the eqns for the E-fields
of different charge distributions contained a factor of
4π. Suggested writing the k in Coulomb’s law to
reflect the 1/4 π and a new constant ε, so that
k=
1
4πε
• ε is called the permittivity (from Latin permittere, to
let go through).
• In a vacuum,
εo = 8.8541878 x 10-12 C2/Nm2
Relative permittivity is ε/ εo
Also called dielectric constant (Ke) and is a unit-less
ratio.
Example 6
• A point charge of 10 micro C is surrounded by water
with a dielectric constant of 80. Calculate the
magnitude of the electric field 20 cm away.
Field Lines
• They diverge away from a + point charge and
converge on a – point charge.
• We call the former a source of the field, the latter the
sink.
• E field lines always begin on the positive charge and
end on the negative charge.
• Lines of the net field never cross
• Dipole: positive and negative charges of the same
magnitude.
Gauss’s Law
• Allows us to calculate the E field of any symmetrical
charge distribution.
• K. F. Gauss. German mathematician and
astronomer (19th C.)
• The product of the area and the strength of the field
passing perpendicularly through it is the flux of the
field.
• Think of it like Continuity Eqn in fluids.
• A1 Eperp1 = A2 Eperp2
• For any closed surface the net flux of the E field
passing through that surface, whatever its shape, is
proportional to the total enclosed charge.
• Gauss’s Law becomes: check board.
• Can’t use calculus, so we only deal in symmetrical
charge distributions.
• Charge density
• Linear charge density:
λ = Q/L
Surface charge density:
σ = Q/A
Volume charge density:
ρ = Q/V
Example 7
• A long, straight wire of length L, in air, carries a
uniform positive charge Q. Use Gauss’s Law to find
the electric field at a perpendicular distance r from
the wire at a point that is far from the conductor’s
ends. To operate in a field region free of the
complications of end effects, we limit the analysis to
the domain where L >> r.
• When a charge is made to move against the influence
of an electric field, it will experience a change in it’s
electrical U.
• Fig 16
• The opposite happens when we allow it to move w/
the field. As speed increases, UE decreases
Electric Potential
• Aka Potential, voltage
• Potential energy per unit charge at any point in
space you want to find.
• Units: J/C
• Volts (V), after Alessandro Volta
Potential = (UE)/q
Example 1
• The most commonly used unit of E field is the volt
per meter. Show that this is consistent with the units
introduced thus far.
• ΔV (potential difference) = Vf – Vi
• ΔV = W/q
Example 2
• A small sphere carrying a + charge of 10 microCoulombs is moved against an E field through a
potential difference of 12.0 V. How much work was
done by the applied force in raising the potential of
the sphere?
• There are no cosmic signposts floating around saying
“HERE BE 0 ELECTRIC U.”
• (sign posts talk pirate style)
• Only differences in potential are important.
• Work done against the E field increases U, work
done by the E field decreases U
• Work done on a charge in moving it from point A
to B is independent of the path taken
• When a particle with charge +qe moves in a field,
dropping 1 V in potential, it will decrease its
electrical U by ΔU= qe ΔV = 1.6 x 10-19 J.
• A lot of stuff on the atomic level involve energies of
about this amount, so we measure this bit of energy
with a new unit: electron volt (eV).
• 1 eV = 1.6 x 10-19 J
Example 3
• Figure 16 shows the basic elements of a CRT.
Electrons are “boiled” off a heated cathode and
emerge through a pinhole, being drawn toward the
first anode, which is at a relatively small positive
potential above the cathode. A second anode that is
20 kV above the cathode accelerates the beam up to
speed. Determine the change in U of each e- . Assume
that each e- has negligible motion at the cathode. Find
its final speed.
• In a uniform E field,
VB – VA = ±𝐸𝑑
• V is + (voltage rise) when the displacement is
opposite to the field and – when it’s same dir’n.
• In a 12 V battery, the + terminal is 12 V higher than
the negative terminal.
Example 4
• If the separation of the plates in Fig 16.7 is 2.00 mm,
determine the magnitude of the E field in the air gap.
• VB – VA = kQ [(1/rB) – (1/rA)]
• V = (kQ)/r
• This is the potential at a distance r from a positive
point-charge Q measured with respect to the zero at
infiniti.
Example 4
• What is the potential difference encountered in going
from close in (point B) to far out (point A) in Fig
16.8, if the tiny sphere carries a charge of 10 microC,
rA = 20 cm and rB = 10 cm?
Conservation of Charge
• Charge is conserved.
• Simplest way to look at it: if charge disappears, then
you have problems with conservation of energy
Capacitors
• Volta coined the term “electrical capacity.” Only in
Italian, I’m assuming.
• At a give potential (V), the amnt of charge (Q) that
can be stored by a body depends on its physical
characteristics, all of which we lump together under
the name capacitance (C)
• C = Q/V
• Unit: C/V and in honor of Faraday, it’s called a farad
(F).
• 1 F = 1 C/V
• To find capacitance of a metal sphere of Radius R,
suppose it’s carry a charge Q.
• V= kQ/R= Q/4εoR and so
• C= Q/V= 4πεoR
Example 7
• Determine the capacitance of an isolated metal sphere
50 cm in diameter and immersed in a vacuum.
Parallel Plate Capacitor
• Ben Franklin among the first.
• Sandwiched insulating glass between metal plates.
• Introducing the second, negatively charged, plate
drops the potential of the first plate, which can be
restored to its original value by further increasing the
charge.
• The second plate considerably increase the ability of
the parallel plate capacitor to store charge at a given
voltage.
• ΔV = Ed
• E = σ/ε= Q/A ε
Where A is the area of each plate and σ = Q/A
ΔV = Ed= Q/(A ε)
C=Q/ ΔV= Q/(Qd/A ε)
So…
C = εA/d