Transcript Electric Potential Energy

Electric Energy and Circuits
Electrostatic Equilibrium
• No net motion of charge is occurring within a
conductor
• Meets the following conditions
▫ Electric field is zero everywhere inside the
conductor
▫ Any excess charge on an isolated conductor
resides entirely on on the conductor’s outer
surface
Electrostatic Equilibrium
▫ The electric field just outside a charged
conductor is perpendicular to the conductor’s
surface
▫ On an irregularly shaped conductor; charge
tends to accumulate where the radius of
curvature of the surface is smallest (i.e. Sharp
points.)
▫ Electrostatic Equilibrium = “Equipotential”
Electric Potential Energy, U
• Electric potential energy –the change in
electrical potential energy of a charge
▫ U = Uf - Ui = -W
▫ The work done by a conservative force (e.g., an electric
force and force of gravity) is equal to the negative of
the change in potential energy.
▫ SI units: Joules (J)
• Similar to energy due to an objects position
above the earth (GPE)
▫ Results from the interaction of two objects’
charges; not mass
𝑊 = −𝑞𝑜 Ed
∆U = -W therefore: ∆U = −𝑞𝑜 Ed
(a) A positive test charge q0,
experiences a downward force due
to the electric field 𝑬. If the charge
is moved upward a distance d, the
work done by the electric field is
–qoEd. At the same time, the
electric potential energy of the
system increases by qoEd. The
situation is analogous to that of an
object in gravitational field. (b) If
a ball is lifted against the force
exerted by gravity, the
gravitational potential energy of
the system increases.
Electric Potential Energy
………final thoughts
• Work is done any time charge moves b/c of an
electric force
W = -∆U (U = potential energy)
• Electric potential energy is conserved
SI = joule (J)
• q = charge
• E = strength of electric field
• d = displacement
▫ Magnitude of displacement’s component in the
direction of the eclectic field.
Electric Potential Energy
• (-) indicates that PEelectric will↑ if (-q) and ↓ if
(+q)
• Valid only in a uniform electric field
• Any displacement perpendiular to and electric
field does not change the PEelectric
Peelectric In a Uniform Electric Field
Toward E
Opposite E
+ charge
- charge
Loses Peelectric Gains Peelectric
Gains Peelectric Loses PEelectric
Electric Potential, V
• Electric Potential: the change in electric
potential:
▫ V = Vf – Vi = U/q0 = (-W)/q0
▫ V is the change in electric potential energy per
charge.
▫ SI units: Joules/Coulomb (J/C) = Volt (V)
• Relationship between U and V:
▫ U = q0 V
▫ Both are scalar quantities.
Electric Potential, V
• Another commonly used unit of energy is the
electron volt (eV):
• 1 eV = (1.60x10-19 C)(1 V) = 1.60x10-19 J
Practice Problem
Exercise 20-1 pg. 664 AP book
Find the change in electric potential energy, U,
as a charge of (a) 2.20 x 10-6 C or (b) -1.10 x 10-6
C moves from a point A to a point B, given that
the change in electric potential between these
points is V = VB – VA = 24.0 V.
Connection between Electric Field and
Electric Potential
• Connection between the electric field and the
electric potential:
▫ E = -V/s
▫ V = -Es
• The electric field depends on : rate of change of the
electric potential with position.
• The electric potential decreases as one moves in the
direction of the electric field.
• SI units for E: 1 N/C = 1 V/m
Practice Problem: Plates at Different
Potentials
Example 20-1 AP book pg. 665
A uniform electric field is
established by connecting
the plates of a parallel-plate
capacitor to a 12-V battery.
(a) If d = 0.75 cm, what is the
magnitude of the electric
field in the capacitor?
(b) A charge of +6.24x10-6 C
moves from the positive
plate to the negative plate.
Find the change in electric
potential energy.
(In electrical systems, we shall assume
that gravity can be ignored, unless
specifically instructed otherwise.)
Energy Conservation

Energy conservation: For a charged object in
an electric field, its total energy must be
conserved.


KA + UA = KB + UB, or
(1/2)mvA2 + UA = (1/2)mvB2 + UB
Practice Problem: From Plate to Plate
Example 20-2 AB book pg. 668
• Suppose a charge q = +6.24 x 10-6 C is released
from rest at the positive plate and that it reaches
the negative plate with a speed of 3.4 m/s.
• (a) What is the mass of the charge?
• (b) What is its final kinetic energy?
The Electric Potential of Point Charges
Exercise 20-2 AP book pg. 670

Electric potential V produced by a
point charge q at a distance r:


Conventionally, choosing the electric
potential to be zero at infinity, V = kq/r
Electric potential energy U for point
charges q and q0 separated by a
distance r:

U = q0V = kq0q/r
The Electric Potential of Point Charges
Exercise 20-2 AP book pg. 670

Find the electric potential by a point
charge of 6.80x10-7 C at a distance of
2.60 m.
“U’ associated with multiple charges
• If a second charge is placed nearby, there will
be Peelectric associated with the two charges.
• Electric field is not uniform
“U” associated with multiple charges
• Point of reference for Peelectric is assumed to
be infinity
▫ Peelectric goes to zero as r btwn q’s goes to
infinity
• Like charges repel – (+) W must be done to
bring them together
▫ Peelectric is (+) for like charges and (-) for
opposite charges
20
Electric Fields in Circuits
• Point away from positive terminal, towards
negative
• Channeled by conductor (wire)
• Electrons flow opposite field lines (neg. charge)
electrons & direction of motion
E
E
E
E
Electric field direction