Electric potential energy

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Electric Potential
We introduced the concept of potential energy in mechanics
Let’s remind to this concept and apply it to introduce electric potential energy
We start by revisit the work done on a particle of mass m
-by a force in general
-by a conservative force such as the gravitational force
In general, work done by a force F moving a particle from point ra to rb
b
Wab   F d r
a
rb (t f )
x2
r a (t0 )
x1
In general we have to specify how we get from a to b, e.g., for friction force
However,
for a conservative force such as gravity we remember
where  is the potential energy
F  
b
With this we obtain Wab  
  d r   d    
a
Gravity as an example
b
b
 a 
 
a
Independent of the path
between r b (t f ) and r a (t0 )
M
Gravitational force derived from  ( r )  G
r
h
Pot. energy
depends
on h, not how
to get there.
Electric potential energy
Can we find a function U =U (r) such that U ( r )  F
is the force exerted by a point charge q on a test charge q0 ?
We expect the answer to be yes, due to the similarity between Coulomb force
and gravitational force
F
gravity
m m0
 G 2 rˆ
r
mm0
 ( r )  G
,    / m0
r
Potential
Potential energy
F
Coulomb
1 qq0

rˆ
2
4 0 r
Let’s try U ( r ) 
In fact we see
F
Coulomb
1 qq0
4 0 r
dU
 U ( r )  
rˆ
dr
1 qq0

rˆ
2
4 0 r
simple because of radial symmetry where U(r)=U(r)
We conclude
Electric potential energy of electrostatically interacting point charges q and q0
1 qq0
U (r) 
4 0 r
U
U
attractive potential qq0<0
repulsive potential qq0>0
r
r
As always, potential defined only up to an arbitrary constant.
Expression above uses U(r)=0 as reference point
We know already the superposition principle for electric fields and forces,
can we find a net potential energy for
F ( r )  F 1 ( r )  F 2 ( r )  F 3 ( r )  ... q0 interacting with several point charges?
Force exerted on q0 by charge q3 at r3
Force exerted on q0 by charge q2 at r2
Force exerted on q0 by charge q1 at r1
Net force q0 experiences
y
 1 q1q0 
F 1 ( r )  U1 ( r )   

4

r

r
1 
0

 1 q2 q0 
F 2 ( r )  U 2 ( r )   

4

r

r
2 
0

...
q1
Note:
textbook on p. 785 defines ri  r i  r
I prefer to keep r-dependence
explicitly visible
r1
r
q0
r2
r3-r
r3
q3
x
q
F ( r )  F 1 ( r )  F 2 ( r )  F 3 ( r )  ...  U1 ( r )  U1 ( r ) 2 U1 ( r )  ...
  U1 ( r )  U 2 ( r )  U 3 ( r )  ...   U ( r )

q0  q1
q2
q3
q0
qi
U (r) 



...



i r  r
4 0  r  r1 r  r 2 r  r 3
4

i
0

The last expression answered the question about the potential energy of the
charge q0 due to interaction with the other point charges q1, q2, …,
q1
y
Those point charges q1, q2, …, interact as well.
Each charge with all other charges
r1
q0
r
r2
r3-r
r3
q3
x
q2
If we ask for the total potential energy of the collection of charges we obtain
U
1
4 0
r
i j
qi q j
i
rj
makes sure that we count each pair only once
This is the energy it takes to bring the charges from infinite separation to
their respective fixed positions ri
Clicker question
2) v  2mEd / q
3) v  qEd / 2m
4) v  qEd / 4 m
0
5) None of the above
-------------------------------------------------
1) v  2qEd / m
+++++++++++++++++++++++++++++++
What is the speed of charge q after moving in the field E from the
positive to the negative plate.
Neglect gravity.
+
d
Electric potential
Goal: Making the potential energy a specific, test charge independent quantity
We are familiar by now with the concept of creating specific quantities, e.g.,
Force on a test charge F  qE
Electric field: test charge independent, specific quantity E  F / q
Gravitational potential energy
 ( r )  G
Mm0
r

M
 G
test mass independent, specific potential  
m0
r
Electric potential V
V
U
q0
Specific, test charge independent potential energy.
The SI unit of the potential is volt (V) .
Meaning of a potential difference
Point a
Wa->b work done by electric force during
displacement of charge q0 from a to b.
Wab  U   Ub  Ua 
 Ub Ua 
Wab
U

 

   Vb  Va   Va  Vb
q0
q0
 q0 q0 
Voltage of the battery
Point b
Alternatively we can ask:
What is the work an external force, F, has to do to move charge q0 from b to a
This force is opposite to the electric force, Fel, above.
Hence:
a
a
b
1
1
1
1
W / q0   Fd r    F el d r   F el d r  Wab  Va  Vb
q0 b
q0 b
q0 a
q0
We know these two alternative interpretations already from mechanics
z
a
Fg=-mg
b
b
Wab   mg dz  mg ( zb  za )    mg ( za  zb )   a   b
a
a
F=mg
To slowly (without adding kinetic energy) move mass from b to a we
need an external force acting against gravity
b
a
W   mg dz  mg ( za  zb )   a   b  
b
Relation between electric potential & electric field
b
From
b
Wab   F d r  q0  E d r
a
a
and
Wab
 Va  Vb
q0
F  q0 E
b
Va  Vb   E d r
a
We obtain the potential difference (voltage)
from the path independent line integral
taken between points a and b
Let’s calculate the potential of a charged conducting sphere
by integrating the E-field
b
We start from Va  Vb 
R
r
point a becomes variable
point in distance r
0
point b becomes reference
point at r

dr
V ( r )  V ( r  ) 
4 0 r r2
0.5
for r>R
:=0
0.0
0
1.0
0
V(r)4 R/Q
a
Q
2
E(r)4 R /Q
1.0
 E dr
1
2
r/R
3
4
Q  1 
V (r) 
 
4 0  r 


r
Q
4 0r
For r<R
R

Q 
dr 

V ( r )  V ( r  ) 
0
dr


R r2 
4 0  r
0.5
V (r) 
0.0
Q
4 0 R
An important application of our “potential of a conducting sphere”- problem
R
According to our considerations above we find at the
surface of the conducting sphere:
Vsurface 
Q
4 0 R
 Esurface R
There is a dielectric breakdown field strength, Em, for all insulating
materials including air
For E>Em air becomes conducting due to discharge
Vm max potential of a sphere before discharge in air sets in
From Em 
R depends on radius
Pinwheel: Electrons from the generator leave the
pinwheel at point of small R. This charge collects on
adjacent air molecules. Electrostatic repulsion
propels the pinwheel
Demonstration: Surface Charge Density
How do we actually measure the charge on the proof plane ?

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