Transcript Electric Potential Energy

Electric Potential Energy
• A charge q in an electric field behaves similarly to
a mass m in a gravitational field.
• The electric force F = qE is conservative.
Work done  Fx cos   qEs cos 
Change in potential energy
from position a to b:
U  U b  U a  qEs cos 
ΔU is independent of the
path from position a to b:
Electric Potential
The electric field is related to how fast the potential is
changing:
For a constant electric field:
Electric Potential and Potential Difference
Ua
Va 
(scalar quantity)
Electric Potential:
q
- independen t of q
Electric potential
difference ΔV
between a and b:
- depends on position a
Ub U a
V  Vb  Va 

q
q
U

  E  x x is along field lines
q
Electric Potential and Potential Energy
U  U b  U a  qVb  Va 
Difference in
potential energy
Difference in
electric potential
• Electric potential V has units of Joules/Coulomb which
is defined as a Volt:
1 Volt = 1 Joule/Coulomb
• One Joule is the work done in moving one Coulomb of
charge through a potential difference of one Volt.
• Electric field has units of Newtons/Coulomb or
Volts/meter:
1N/C = 1 J/(mC) = 1 V/m
Analogy between Electric and
Gravitational Fields
q
E
m
G
d
q
m
ΔU = qΔV = - qEd
ΔU = - mgd
The charge and mass lose potential energy and gain
kinetic energy when they move in the direction of the
field.
Kinetic Energy of a Charge
Accelerated by an Electric Field
• The kinetic energy acquired by an electron or a
proton accelerated through a potential difference
of 1000 Volts:
• Uba = qVba = (1.60 x 10-19 C)(1000 V)
= 1.60 x 10-13 J = 1000 eV (electron volts)
= 1 keV (kilo electron volt)
 One electron-volt (1 eV) is the kinetic energy
gained by an elemental charge (electron or proton)
when it is accelerated through a potential
difference of one Volt.
1 eV = 1.6 x 10-19 J
Electric Potential Energy:
The Electron Volt
• Suppose a point charge q is moved between two points
a and b in space, where the electric potentials due to
other charges are Va and Vb.
• The change in potential energy is:
ΔU = Ub – Ua = q(Vb – Va) = qVba
• Unit = Electron Volt (eV):
1 eV = 1.6 x 10-19 J
• e.g. a proton accelerated through a potential
difference of 200 kV acquires a kinetic energy of
• 200 keV (losing 200 keV of electric potential energy).
Electric Potential due to Point Charge
Electric Field:
Q
E  k 2 (radially outward)
r
Using Calculus, it can be shown
1 1
V  Vb  Va  kQ   
 rb ra 
Convention: V=0 at infinite r
Q
V  k (electric potential)
r
Electric Potential due to a Point Charge
Electric potential at a
distance r from a positive
charge Q
Electric potential at a
distance r from a negative
charge Q
Equipotential Surfaces
• Equipotential surfaces are surfaces of constant electric
potential (just as lines of constant elevation on a topological
map are lines of constant gravitational potential).
• Equipotential surfaces are always perpendicular to the
direction of the electric field. (just as the “fall line” is
perpendicular to the contour lines on a topological map).
Charged Parallel Plates
Two Equal and Opposite Charges