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Unformatted text preview: • Physics 227: Lecture 5
Gauss’s Law Leftovers, Work /
Potential Energy in Electrostatics
Lecture 4 review: • Gauss’s Law: for a closed surface, Φ = ∫ E.dA = ∫ E⊥dA = ∫ E
cosφ dA = qenclosed/ε0. •
•
• The ﬂux does not depend on the size or shape of the
Gaussian surface, or the position of the charges, it depends
only on the total enclosed charge.
Examples: point charge, inside uniform sphere, inside
spherical shell, inﬁnitely long line charge, inﬁnitely long
cylinder of charge, inside the cylinder of charge, inﬁnite
plane of charge, two planes of charge, ... Conductors: electrons move around so that the ﬁeld inside a
conductor vanishes. Thus requires all net charge is on the
surface. Monday, September 19, 2011 Some Geometry
r dθ ˆ
n dφ dθ r dφ r • The area corresponding to an angle range dθ x dφ is r2 dθ dφ,
if the surface is perpendicular to r vector. • If the surface is tilted by an angle α  ˆ os α = r.n  its actual
cˆ
area in the range of angles dθ x dφ increases by a factor 1/cosα. • With charge density λ, the total charge on the surface is:
dq = rdθ rdφ λ / cosα.
Monday, September 19, 2011 • Field Inside Spherical Shell For an spherical or inﬁnite cylindrical
shell of charge, the ﬁeld inside is 0. • Here we show this occurs because the
ﬁelds from opposite parts of the sphere or
the shell cancel. • What is the ﬁeld at the diamond a
distance z above the center of the circle a 2d slice through a cylinder or sphere. θ
z
α α
r0 r0 • To get the ﬁeld, we need to integrate
o ver angles: ``E = ∫kdq/r2’’. The geometry is
shown for a particular angle θ. (Angles are
measured from (0,z), not the origin.) • The ﬁeld from a small area at the upper right, a distance r away from (0,z),
w ill be kdq/r2 = k (rdθ rdφ λ / cosα) /r2 = k (dθ dφ λ / cosα). This is
independent of r! And independent of α for opposite points from (0,z)! • The ﬁeld from the opposite direction is exactly the same magnitude, so the
two cancel! This only works for the circular geometry  for, e.g., a cube the
factors of cosα would generally be different.
Monday, September 19, 2011 Field at Surface of a Charged Conductor
• What is the ﬁeld at the surface
of a charged conductor? • Draw a small Gaussian box
across the surface. The box should
be small enough that the surface
is ﬂat and the charge density is
constant, to a good approximation,
and should be  / ⊥ to the local
surface. • Gauss’s Law: Φ = ∫E.dA = 4.0.A + 0.A + E.A = EA
• Φ = q/ε0 = σA/ε0
• ➮ E = σ/ε0
• This is a factor of 2 larger than for a plane of charge,
because all the ﬁeld is to one side.
Monday, September 19, 2011 E Field in a Idealized Parallel Plate Capacitor
• (Superposition) Adding the ﬁelds
from two inﬁnite planes of charge
yields the idealized result. • Or we can use Gaussian cylinders
+ properties of conductors +
symmetry: E = E0ŷ, but E = 0 at
the end of the Gauusian cylinder
inside the conductor, and E⊥ = 0 for
sides of the Gaussian cylinders. •
•
• For S2 and S3, qenclosed = 0 ➮ Φ = 0 ➮ Eoutside.A = 0 ➮ E = 0.
For S1, qenclosed = σA ➮ Φ = σA/ε0 ➮ Einside.A = σA/ε0 ➮ Einside = σ/ε0 ŷ. For S4, qenclosed = σA ➮ Φ = σA/ε0 ➮ Einside.A = σA/ε0 ➮ Einside = σ/ε0 ŷ.
(Note: here the normal is in the ŷ direction, so the negative signs cancel.) • Do not get confused  we have ``3’’ ways of getting the same result. Do not
a dd them! Even though there is ±σ, you can get the ﬁeld using one of them!
The other is already included in your work in using Eoutside, in conductor = 0.
Monday, September 19, 2011 iClicker: What is the Surface Charge on a Conductor
Consider a conducting sphere,
w ith a cavity inside.
5 μC We insert a 5 μC charge into
the cavity.
What is the charge on the
inner and outer surfaces of
the conductor?
A. 0 μC, 0 μC.
B. 5 μC, 5 μC.
C. 5 μC, 5 μC.
D. 5 μC, 5 μC.
E. 5 μC, 5 μC. Monday, September 19, 2011 iClicker: What is the Surface Charge on a Conductor
Consider a conducting sphere,
w ith a cavity inside.
5 μC We insert a 5 μC charge into
the cavity.
What is the charge on the
inner and outer surfaces of
the conductor? We need 5 μC on the inside surface to stop
the ﬁeld lines from the +5 μC charge in the
cavity. With a neutral conductor, that requires
we have +5 μC on the outer surface. If the
conductor had total charge Q μC, we would
have Q+5 μC on the outer surface. A. 0 μC, 0 μC.
B. 5 μC, 5 μC.
C. 5 μC, 5 μC.
D. 5 μC, 5 μC.
E. 5 μC, 5 μC. Monday, September 19, 2011 Conservative Forces • Gravity and Electric Forces
are conservative forces. •
•
• You can relate the force to
a potential / potential
energy.
Energy is converted from
potential to kinetic. No
energy is lost.
Changes in the potential /
potential energy, depend
only on the initial and ﬁnal
points of a trajectory, not
on the intermediate path
taken. Monday, September 19, 2011 Force / Work / Potential Energy / Field /
Potential • Let’s review these concepts from gravity ﬁrst  for a mass m in a
constant gravitational ﬁeld, like near the surface of the earth
Fg = −mg z
ˆ Force of gravity integrate differentiate z
F · d = −mg ∆z W= ∆U = − Work done by gravity z
F · d = mg ∆z Potential energy from gravity
often used as U = mgz divide by m
= Fg /m = −g z
g
ˆ
Gravitational ﬁeld Monday, September 19, 2011 integrate
differentiate ∆V = − divide by m · d = g ∆z
gz Just as the ﬁeld = the force / m, the
potential = the potential energy / m Force / Work / Potential Energy / Field /
Potential • Now we move to a charge q in a constant electric ﬁeld, as with an
inﬁnite plane of charge. σ
z
ˆ
Fc = q E = −q
20
Electric Force integrate
differentiate W=
Fc · d = −qE ∆z
z Work done by electric force ∆U = −
Fc · d = qE ∆z
z Potential energy from
electric force often used as U = qEz divide by q σ
z
ˆ
E = F /q = −
20
Electric ﬁeld Monday, September 19, 2011 integrate
differentiate ∆V = − divide by q z
E · d = E ∆z Just as the ﬁeld = the force / q, the
potential = the potential energy / q Force / Work / Potential Energy / Field /
Potential • Now we move to a charge q in a constant electric ﬁeld, as with an
inﬁnite plane of charge. σ
z
ˆ
Fc = q E = −q
Electric Force in 20
units of Newtons integrate
differentiate W=
Fc · d = −qE ∆z
z Work in units of Joules
∆U = − Fc · d = qE ∆z
z Energy in units of Joules
(or “electron Volts”) U = qEz divide by q σ
z
ˆ
E = F /q = −
20 Electric ﬁeld units: Volts/m
or Newtons/Coulomb Monday, September 19, 2011 integrate
differentiate ∆V = − divide by q z
E · d = E ∆z Potential V in units of Volts = J/C iClicker: Work and Potential energy
5 μC v E Note: in all those formulas you have
to watch your limits of integration /
signs. Let’s see about your intuition!
The ﬁeld points left.
The 5 μC charge moves right.
Which of the following is correct? A. The ﬁeld does positive work on the charge and potential energy increases. B. The ﬁeld does positive work on the charge and potential energy decreases.
C. The ﬁeld does negative work on the charge and potential energy increases.
D. The ﬁeld does negative work on the charge and potential energy decreases.
E. The ﬁeld does no work on the charge and potential energy increases.
Monday, September 19, 2011 iClicker: Work and Potential energy
5 μC v E E points “downhill”, so the
potential energy for q>0
increases. Going “uphill” slows
the charge down, so the work
done by the ﬁeld is negative. Note: in all those formulas you have
to watch your limits of integration /
signs. Let’s see about your intuition!
The ﬁeld points left.
The 5 μC charge moves right.
Which of the following is correct? A. The ﬁeld does positive work on the charge and potential energy increases. B. The ﬁeld does positive work on the charge and potential energy decreases.
C. The ﬁeld does negative work on the charge and potential energy increases.
D. The ﬁeld does negative work on the charge and potential energy decreases.
E. The ﬁeld does no work on the charge and potential energy increases.
Monday, September 19, 2011 Force / Work / Potential Energy / Field /
Potential for Two Point Charges • Let’s put a charge Q at the origin, generating an E ﬁeld, and move
a charge q around. W=
Fon q kQq
= qE = 2 r
ˆ
r Electric Force integrate
differentiate
F = −∇U
divide by q integrate
Ef rom Q b a ∆U = − c · d = ( kQq − kQq )
F
l
ra
rb
Work done by electric force b a c · d = ( kQq − kQq )
F
l
rb
ra Set Potential energy = 0 at ∞
a c · −d = kQq
U =−
F
r
ra
∞
divide by q
b kQ kQ
l
E · d = (
−
)
rb
ra
∆V = −
Fon q
kQ
=
= 2 r differentiate
ˆ
a
q
r
Just as the ﬁeld = the force / q, the Electric ﬁeld
Monday, September 19, 2011
E = −∇V potential = the potential energy / q More than Two Point Charges
• To get the total potential energy of a charge q0 interacting
w ith several other charges, add up the several potential
energies
n
n
kq0 qi
qi
U0 =
= kq0
r0 i
r0 i
i=1,...
i=1,... • To get the total potential energy of the entire system of
charges, add up the potential energies of all the pairs of
charges U0 = kqi qj
i<j rij We need i<j, or we will count each pair twice, or even get ∞’s from ri2/0!
Monday, September 19, 2011 Potential Energy of a 3charge System
q
A charge q is brought from ∞ so that it
and two ﬁxed charges, of ±q, form an
equilateral triangle.
What is the potential energy of the third
charge? What is the potential energy of
the system? q A. U3 = 2kq2/d, Usys=6kq2/d. q q B. U3 = 0, Usys=kq2/d.
C. U3 = 2kq2/d, Usys=3kq2/d.
D. U3 = 0, Usys=kq2/d.
E. U3 = 0, Usys=2kq2/d. Monday, September 19, 2011 Potential Energy of a 3charge System
q
A charge q is brought from ∞ so that it
and two ﬁxed charges, of ±q, form an
equilateral triangle.
What is the potential energy of the third
charge? What is the potential energy of
the system? q A. U3 = 2kq2/d, Usys=6kq2/d. q q
Of the 3 pairs, 2 are +w ith U<0, and 1 is  with
U>0, so Utotal = kq2/d. The
3rd charge interacts with +
and , so U3 = 0.
Monday, September 19, 2011 B. U3 = 0, Usys=kq2/d.
C. U3 = 2kq2/d, Usys=3kq2/d.
D. U3 = 0, Usys=kq2/d.
E. U3 = 0, Usys=2kq2/d. Potential Energy of a 3charge System
q Do these two systems have the same, or opposite,
system potential energies?
One is +q, q, q, the other is +q, +q, q.
q
q q q In both cases, we have 2
+ pairs with U<0, and 1
same sign pair U>0, so Utotal
= kq2/d is the same.
Monday, September 19, 2011 q q q Thank you, and
See you next Thursday Monday, September 19, 2011 ...
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 Fall '11
 RonaldGilman
 Physics, Electrostatics, Energy, Potential Energy, Work

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