Boundary conditions example
Infinite plane with
charge density σ
+σ
Is the E field continuous
across the plane ?
No !
Is the electric potential V
continuous across the plane ?
Yes ! (energy is not
created or destroyed)
Conclusion is general
Hints for 23.49
+q
Assume the
potential at
infinity is zero
rb
-q
ra
A metal sphere with
radius r is supported on
an insulating stand in
the center of a hollow
metal spherical shell of
radius b.
There is a charge –q
on the inner sphere
and a charge +q on
the outer shell.
One approach: find the E fields and then the
potentials. Match the boundary conditions
Hints for 23.49 cont’d
Make a Gaussian surface inside the
sphere. What is E inside ?
+q
rb
-q
ra
What is the electric potential for r<ra ?
V=V0 we will determine this constant later.
Hints for 23.49 (cont’d)
Now make a spherical Gaussian surface at radius r
+q
r
-q
rb
ra
What is the charge enclosed ?
What is the E field ?
−q
E=
2
4πε 0 r
V (r ) =
−q
4πε 0 r
+ V1
Hints for 23.49 (cont’d)
Now make a spherical Gaussian surface at radius r
outside.
+q
r
rb
-q
ra
What is the charge enclosed ?
What is the E field ?
What is the potential ?
V (r ) = V2 = 0
Hints for 23.49 (cont’d)
Now put all the electric potentials together.
+q
rb
-q
ra
What happens at
the boundaries ?
V (r ) = V0
−
q
−
q
=
;
=
=
+
r
r
V
V
V
a
0
1
r <r<r V ( r ) =
+ V1
πε
4
r
0
a
4πε 0 r
−q
+ V1
r = rb ;V = 0 =
r>r
V (r ) = V2 = 0
4πε 0 rb
r<r
a
a
b
b
Hints for 23.49 (cont’d)
r = rb ;V = 0 =
−q
+ V1
4πε 0 rb
−q
+ V1
r = ra ;V = V0 =
4πε 0 ra
r<a
a<r<b
r>b
V1 =
q
4πε 0 rb
−q
q
+
V0 =
4πε 0 ra 4πε 0 rb
Calculate Vab=Vb-Va
V (r ) = V0
−q
+ V1 V = q ⎛ 1 − 1 ⎞ = V
V (r ) =
⎜
⎟ 0
ab
4πε 0 r
4πε 0 ⎝ ra rb ⎠
V (r ) = V2 = 0
Hints for 23.87
Nuclear
Fission
Q=96e
U236
Q=46e
4π 3
V /2=
Rdaughter
3
R=7.4 x 10-15 m
Q=46e
Half the volume
and half the
charge of the
original nucleus
What is the kinetic energy of each
daughter nucleus when they are
far apart ?
Hints for 23.87
Q=96e
R=7.4 x 10-15 m
U236
What is the kinetic energy of each
daughter nucleus when they are
far apart ?
Q=46e
Rdaughter Rdaughte
2
k (46e)
U=
2 Rdaughter
U
K=
2
Q=46e
Numerical final
result: 253 kilotons
of TNT from 10 kg
of uranium !!
This is “electric potential energy”