Electric Potential Energy of the System of Charges
Potential energy of a test charge q0
in the presence of other charges
U
q0
qi

4 0 i r i
Potential energy of the system of charges
(energy required to assembly them together)
U
1

qi q j
4 0 i  j r ij
Potential energy difference can be equivalently described as a work
done by external force required to move charges into the certain
geometry (closer or farther apart).


External force now is opposite to Wa b  (U b  U a )   Fext d l
the electrostatic force
Electric Potential Energy of System
• The potential energy of a system of two point charges
q1q2
U  ke
r12
• If more than two charges are present, sum the energies of every
pair of two charges that are present to get the total potential energy

U total  ke 
i, j
qi q j
rij
 q1q2 q1q3 q2 q3 
U total  ke 



r13
r23 
 r12
Electric potential is electric potential energy per unit charge
Finding potential (a scalar) is often much easier than the field
(which is a vector). Afterwards, we can find field from a potential
U
V
q0
Units of potential are Volts [V]
1 Volt=1Joule/Coulomb
If an electric charge is moved by the electric
field, the work done by the field
Wa b
U

 (Va  Vb )
q0
q0
Potential difference if often called voltage
Two equivalent interpretations of voltage:
1.Vab is the potential of a with respect to b, equals the work done
by the electric force when a UNIT charge moves from a to b.
2. Vab is the potential of a with respect to b, equals the work that must
be done to move a UNIT charge slowly from b to a against the
electric force.
Potential due to the point charges
1
dq
V
4 0  r
Potential due to a continuous
distribution of charge
Finding Electric Potential through Electric Field
b

Wa b
 Va  Vb   E d l
q0
a

Some Useful Electric Potentials
• For a uniform electric field
r r
r
V    E  dl   E 
• For a point charge
q
V  ke
r
• For a series of point charges
qi
V  ke 
ri
r
r r
 dl  E  l
Potential of a point charge
Moving along the E-field lines means moving in the direction of
decreasing V.
As a charge is moved by the field, it loses it potential energy, whereas
if the charge is moved by the external forces against the E-field, it
acquires potential energy
• Negative charges are a potential minimum
• Positive charges are a potential maximum
Positive Electric Charge Facts
• For a positive source charge
– Electric field points away from a positive source charge
– Electric potential is a maximum
– A positive object charge gains potential energy as it moves
toward the source
– A negative object charge loses potential energy as it moves
toward the source
Negative Electric Charge Facts
• For a negative source charge
– Electric field points toward a negative source charge
– Electric potential is a minimum
– A positive object charge loses potential energy as it moves
toward the source
– A negative object charge gains potential energy as it moves
toward the source
Electron Volts
Electron volts – units of energy
U  eVab
1 eV – energy a positron (charge +e) receives when it goes
through the potential difference Vab =1 V
Unit: 1 Volt= 1 Joule/Coulomb (V=J/C)
Field: N/C=V/m
1 eV= 1.6 x 10-19 J
Just as the electric field is the electric force
per unit charge, the electrostatic potential is
the potential energy per unit charge.
Examples
A small particle has a charge -5.0 mC and mass 2*10-4 kg. It moves
from point A, where the electric potential is fa =200 V and its speed
is V0=5 m/s, to point B, where electric potential is fb =800 V. What
is the speed at point B? Is it moving faster or slower at B than at A?
E
2
2
A
B
F
mV0
mV
 qa 
 qb
2
2
Vb ~ 7.4 m / s
In Bohr’s model of a hydrogen atom, an electron is considered
moving around a stationary proton in a circle of radius r. Find
electron’s speed; obtain expression for electron’s energy; find total
energy.
11
e2
V2
r

5.3

10
m
U
Fe  ke 2  m
K
r
r
2
T  13.6 eV
T  K U
Calculating Potential from E field
• To calculate potential function from E field
V   
f
i
r r
E  ds

   (E x iˆ  E y ˆj  E z kˆ )  dxiˆ  dyˆj  dzkˆ
f
i


f
i
E x dx  E y dy  E z dz

When calculating potential due to charge distribution, we calculate potential explicitly
if the exact distribution is known.
If we know the electric field as a function of position, we integrate the field.
b

   E d l
a
Generally, in electrostatics it is easier to calculate a potential (scalar) and then find
electric field (vector). In certain situation, Gauss’s law and symmetry consideration
allow for direct field calculations.
Moreover, if applicable, use energy approach
rather than calculating forces directly
(dynamic approach)
Example: Solid conducting sphere
Outside: Potential of the point charge
1
q
V
4 0 r
Inside: E=0, V=const
Potential of Charged Isolated Conductor
• The excess charge on an isolated conductor will distribute itself
so all points of the conductor are the same potential (inside and
surface).
• The surface charge density (and E) is high where the radius of
curvature is small and the surface is convex
• At sharp points or edges  (and thus external E) may reach high
values.
• The potential in a cavity in a conductor is the same as the
potential throughout the conductor and its surface
At the sharp tip (r tends to zero), large
electric field is present even for small
charges.
Lightning rod – has blunt
end to allow larger charge
Corona – glow of air due to gas discharge built-up – higher probability
near the sharp tip. Voltage breakdown of of a lightning strike
the air
Vmax  3  10
6
V /m
Vmax  REmax
Example: Potential between oppositely charged parallel plates
From our previous examples
U ( y )  q0 Ey
V ( y )  Ey
Vab
E
d
Easy way to calculate surface
charge density

 0Vab
d
Remember! Zero potential doesn’t mean the conducting object has no
charge! We can assign zero potential to any place, only difference in
potential makes physical sense
Calculating E field from Potential
• Remembering E is perpendicular to equipotential surfaces
E  V
 V ˆ V ˆ V ˆ 
E  
i
j
k
y
z 
 x
V
V
V
Ex  
 Ey  
 Ez  
x
y
z
Example: Charged wire
We already know E-field around the wire
only has a radial component
b
1 
rb

ln
Er 
;    E  dr 
2 0 ra
2 0 r
a
Vb = 0 – not a good choice as it follows
Va  
Why so?
We would want to set Vb = 0 at
some distance r0 from the wire
r - some distance from the wire
r0

V
ln
2 0 r
Example: Sphere, uniformly charged inside through volume
r
q  Q 
R
R
3
'
E
Q - volume density of charge

V

r
3 0
R
 ( r  R)
r
 R  r   E dr
2
R
keQ r r
R    3
|R
R 2
keQ
R 
R
Q - total charge
keQ 
r2 
r 
3  2 
2R 
R 
This is given that at infinity  0