Ben Gurion University of the Negev
www.bgu.ac.il/atomchip
Physics 2B for Materials and Structural Engineering
Lecturer: Daniel Rohrlich
Teaching Assistants: Oren Rosenblatt, Shai Inbar
Week 4. Potential, capacitance and capacitors – E from V •
equipotential surfaces • E in and on a conductor • capacitors and
capacitance • capacitors in series and in parallel
Source: Halliday, Resnick and Krane, 5th Edition, Chaps. 28, 30.
With 100,000 V on her body, why is this girl smiling???
E from V
Let U(r2) be the potential energy of a charge q at the point r2.
We found that it is minus the work done by the electric force
Fq in bringing the charge to r2:
r2

U (r2 )  U (r1 )   Fq (r)  dr .
r1
If we divide both sides by q, we get (on the left side) potential
instead of potential energy, and (on the right side) the electric
field instead of the electric force:
r2

V (r2 )  V (r1 )   E(r)  dr .
r1
E from V
Let V(r2) be the electric potential at the point r2. We have just
obtained it from the electric field E:
r2

V (r2 )  V (r1 )   E(r)  dr .
r1
Now let’s see how to obtain E from V. For r2 close to r1 we
can write r2 = r1 + Δr and expand V(r2) in a Taylor series:
V (r2 )  V (r1  r )
V
V
 V (r1 ) 
x 
x x
y
1
y1
V
y 
z
z  ... ;
z1
E from V
We can write this expansion more compactly using the “del”
(gradient) operator:
    
   , ,  ,
 x y z 
so
 V V V 
 ,
V (r1 )  
,
,
 x y z 
with all the derivatives evaluated at the point r1. We also write
V
V
V
V (r1 )  r 
x 
y 
z .
x
y
z
E from V
So now we can write
V
V
V
V (r1  r )  V (r1 ) 
x 
y 
z  ...
x
y
z
 V (r1 )  V (r1 )  r  ...
and therefore
r2

V (r2 )  V (r1 )   E(r )  dr
r1
V (r1  r )  V (r1 )  E(r1 )  r
V (r1 )  r  E(r1 )  r .
E from V
The equation
V (r1 )  r  E(r1 )  r
is a scalar equation, but since we can vary each component of
Δr independently, it actually yields three equations:
V (r1 )x   E x (r1 )
,
V (r1 )y   E y (r1 )
,
V (r1 )z   E z (r1 )
,
E from V
The equation
V (r1 )  r  E(r1 )  r
is a scalar equation, but since we can vary each component of
Δr independently, it actually yields three equations, i.e.

V (r1 )   E x (r1 ) ,
x

V (r1 )   E y (r1 ) ,
y

V (r1 )   E z (r1 ) ,
z
E from V
which reduce to a single vector equation:
E(r)  V (r) .

V (r1 )   E x (r1 ) ,
x

V (r1 )   E y (r1 ) ,
y

V (r1 )   E z (r1 ) ,
z
E from V
Example 1 (Halliday, Resnick and Krane, 5th Edition, Chap. 28,
Exercise 34): Rutherford discovered, 99 years ago, that an
atom has a positive nucleus with a radius about 105 times
smaller than the radius R of the atom. He modeled the electric
potential inside the atom (r < R) as follows:
Ze  1 3
r 2 
V (r ) 


,
2
4 0  r 2R 2R 
where Z is the atomic number (number of protons). What is the
corresponding electric field?
E from V
Ze  1 3
r 2 
Answer: E(r )  V (r )  


4 0  r 2 R 2 R 2 
Ze  1
r 



r
4 0  r 2 R 2 
Ze  1
r 



rˆ ,
4 0  r 2 R 2 
since
r
r
r xˆ x  yˆ y  zˆ z r
r  xˆ
 yˆ
 zˆ

  rˆ .
x
y
z
r
r
E from V
Example 2: Electric field of a dipole
We found that the electric potential V(x,y,z) of a dipole made of
charges q at (0,0,d/2) and –q at (0,0,–d/2) is
1
1
V ( x, y , z ) 
   ,
4 0  r r 
z
d/2
q
r–
(x,y,z)
r+
–d/2
d 2
where r  x  y  ( z  )
2
2
2
.
E from V
V
qx
E x (r )  

x 4 0
 1
1 


 ,
3
3
r  
 r 
V
qy
E y (r )  

y 4 0
 1
1 


 ,
3
3
r  
 r 
V
q
E z (r )  

z 8 0
 2z  d 2z  d 


 ,
3
3
r  
 r 
z
d/2
r–
(x,y,z)
r+
–d/2
d 2
where r  x  y  ( z  )
2
2
2
.
E from V
Now we will consider the case d << r– , r+ and use these rules
for 0 < α << 1 (derived from Taylor or binomial expansions) to
approximate E:
1  1

2
 ... ,
1
 1    ... ,
1
(1   ) n  1  n  ... .
E from V
For d << r– , r+ we have
2
d
2
2 
r  x  y   z   
2

1/2
 zd d 
 r 1 


2
2
4r 
 r
2
2
d
x 2  y 2  z 2  zd 
4
, where r  x 2  y 2  z 2
z
d/2
r–
(x,y,z)
–d/2
r+
.
E from V
For d << r– , r+ we have
r  r 1   1/2 , where   
1
(r ) 3


1   3/2
3
1
r
zd
r
2
d2

4r
r–
(x,y,z)
–d/2
r+
, so
1  3  1  3zd 3d 2 

1   
1  2  2  .
3 
3
r  2  r  2r
8r 
z
d/2
2
E from V
1  3zd 3d 2  1
So


1  2  2   3
3
3
3
r  r  r  2r 8r  r
1
1

3zd
r
Ex 
3xzqd
4 0 r
5
5
, Ey 
; and now
3 yzqd
4 0 r
5
, Ez 
z
d/2
r–
(x,y,z)
–d/2
r+
 3zd 3d 2 
1  2  2 
8r 
 2r
(3z 2  r 2 )qd
4 0 r
5
.
E from V
Ex 
3xzqd
4 0 r 5
, Ey 
3 yzqd
4 0 r 5
, Ez 
(3z 2  r 2 )qd
4 0 r 5
We can check that these results coincide with the results we
obtained for the special cases x = 0 = y and z = 0:
Ez 
qd
2 0 z 3
, Ez  
qd
4 0 r 3
z
d/2
r–
(x,y,z)
–d/2
r+
.
.
E from V
Ex 
3xzqd
4 0 r 5
, Ey 
3 yzqd
4 0 r 5
, Ez 
(3z 2  r 2 )qd
4 0 r 5
This may look complicated but it is easier than calculating Ex,
Ey and Ez directly!
z
d/2
r–
(x,y,z)
–d/2
r+
.
Equipotential surfaces
We have already seen equipotential surfaces in pictures of lines
of force:
and in connection with potential energy:
r2
F(r)
r
r1
Equipotential surfaces
All points on an equipotential surface are at the same electric
potential.
Electric field lines and equipotential surfaces meet at
right angles. Why?
F(r)
r
r1
r2
Equipotential surfaces
The surface of a conductor at electrostatic equilibrium – when
all the charges in the conductor are at rest – is an equipotential
surface, even if the conductor is charged:
Small pieces of
thread (in oil)
align with the
electric field due
to two conductors,
one pointed and
one flat, carrying
opposite charges.
[From Halliday,
Resnick and Krane]
Equipotential surfaces
We can also visualize the topography of the electric potential
from the side (left) and from above (right) with equipotentials
as horizontal
V
curves.
y
x
Equipotential surfaces
Quick quiz: In three space dimensions, rank the potential
differences V(A) – V(B), V(B) – V(C), V(C) – V(D) and
V(D) – V(E).
B
9V
A
8V
7V
6V
C
E
D
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
Explanation: A conductor contains free charges (electrons) that
move in response to an electric field; at electrostatic
equilibrium, then, the electric field inside a conductor must
vanish.
Example: An infinite conducting sheet in
a constant electric field develops surface
charges that cancel the electric field inside
the conductor.
+
+
+
+
+
+
+
-
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
Explanation: A charge inside the conductor would imply,
via Gauss’s law, that the electric field could not be zero
everywhere on a small surface enclosing it, contradicting 1.
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
3. The electric field on the surface of a conductor must be
normal to the surface and equal to σ/ε0, where σ is the surface
charge density.
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
3. The electric field on the surface of a conductor must be
normal to the surface and equal to σ/ε0, where σ is the surface
charge density.
Explanation: A component of E parallel to the surface would
move free charges around. (Hence the surface of a conductor
at electrostatic equilibrium is an equipotential surface, even if
the conductor is charged.)
E in and on a conductor
Let’s compare the electric fields due to two identical surface
charge densities σ, one on a conductor with all the excess
charge on one side (e.g. the outside of a charged sphere) and
the other on a thin insulating sheet.
The figure shows a short “Gaussian can” straddling the thin
charged sheet. If the can is short, we need to consider only the
electric flux through the top and bottom.
Gauss’s law gives 2EA = ФE = σA/ε0, so
E = σ/2ε0. But if there is flux only through
the top of the can, Gauss’s law gives
EA = ФE = σA/ε0 and E = σ/ε0.
“Gaussian can”
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
3. The electric field on the surface of a conductor must be
normal to the surface and equal to σ/ε0, where σ is the surface
charge density.
4. On an irregularly shaped conductor, the surface charge
density is largest where the curvature of the surface is largest.
E in and on a conductor
Explanation: If we integrate E·dr along any electric field line,
starting from the conductor and ending at infinity, we must get
the same result, because the conductor and infinity are both
equipotentials. But E drops more quickly from a sharp point
or edge than from a smooth surface – as we have seen, E drops
as 1/r2 from a point charge, as 1/r near a charged line, and
scarcely drops near a charged surface. The integral far from
the conductor is similar for all electric field lines; near the
conductor, if E·dr drops more quickly from a sharp point or
edge, it must be that E starts out larger there. Then, since E is
proportional to the surface charge σ, it must be that σ, too, is
larger at a sharp point or edge of a conductor.
Capacitors and capacitance
A capacitor is any pair of
isolated conductors. We
call the capacitor charged
when one conductor has
total charge q and the other
has total charge –q. (But
the capacitor is then actually
neutral.)
Whatever the two conductors
look like, the symbol for a
capacitor is two parallel lines.
capacitor
battery
switch
Capacitors and capacitance
Here is a circuit diagram with a battery to charge a capacitor,
and a switch to open and close the circuit.
capacitor
battery
switch
Capacitors and capacitance
The charge q on the capacitor (i.e. on one of the two conductors)
is directly proportional to the potential difference ΔV across the
battery terminals: q = C ΔV.
capacitor
battery
switch
Capacitors and capacitance
The charge q on the capacitor (i.e. on one of the two conductors)
is directly proportional to the potential difference ΔV across the
battery terminals: q = C ΔV. (Note that q and ΔV both scale the
same way as the electric field.)
Capacitors and capacitance
The charge q on the capacitor (i.e. on one of the two conductors)
is directly proportional to the potential difference ΔV across the
battery terminals: q = C ΔV. (Note that q and ΔV both scale the
same way as the electric field.)
The constant C is called the capacitance of the capacitor. The
unit of capacitance is the farad F, which equals Coulombs per
volt: F = C/V.
Capacitors and capacitance
Example 1: Parallel-plate capacitor
If we can neglect fringing effects (that is, if we can take the
area A of the conducting plates to be much larger than the
distance d between the plates) then E = σ/ε0 where σ = q/A and
ΔV = Ed = qd/ε0A. By definition, q = C ΔV, hence C = ε0A/d
is the capacitance of an ideal parallel-plate capacitor.
Capacitors and capacitance
Example 2: Cylindrical capacitor
Again, if we can neglect fringing effects (that is, if we can take
the length L of the capacitor to be much larger than the inside
radius b of the outer tube) then
1
q
E (r ) 
,
2 0 Lr
V 
q
2 0 L
b

a
a  r  b ,
dr
q

ln (b/a) ,
r
2 0 L
a
b
2 0 L
C
ln b  ln a
.
Capacitors and capacitance
Example 3: Spherical capacitor
The diagram is unchanged, only E(r) is different:
E (r ) 
1
q
4 0 r 2
b
,
a  r  b ,
q 1 1
V 

   ,
2
4 0 r
4 0  a b 
a
q

dr
4 0 ab
C
.
ba
a
b
Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances.
Here are two capacitors in series:
capacitor
battery
switch
Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances.
Here are two capacitors in series:
Since the potential across both
capacitors is ΔV, we must have
q1 = C1ΔV1 and q2 = C2ΔV2
where ΔV1 + ΔV2 = ΔV. But the
charge on one capacitor comes
from the other, hence q1 = q2 = q.
Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances.
Here are two capacitors in series:
Since the potential across both
capacitors is ΔV, we must have
q = C1ΔV1 and q = C2ΔV2 where
ΔV1 + ΔV2 = ΔV. If the effective
capacitance is Ceff, then we have
q
q
q
 V  V1  V2 

,
Ceff
C1 C 2
1
1
1


.
hence
Ceff
C1 C 2
Capacitors in series and in parallel
For n capacitors in series, the generalized rule is
1
1
1
1


 ... 
Ceff
C1 C 2
Cn
C1 C2
Cn
.
Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances.
Here are two capacitors in parallel:
capacitor
battery
switch
Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances.
Here are two capacitors in parallel:
Since the potential across each
capacitor is still ΔV, the charge
on the capacitors is q1 = C1ΔV
and q2 = C2ΔV. If the effective
capacitance is Ceff, then we have
C1ΔV + C2ΔV = q1 + q2 = CeffΔV,
thus capacitances in parallel add:
C1 + C2 = Ceff .
Capacitors in series and in parallel
For n capacitors in parallel, the generalized rule is
C1
C2
Cn
Ceff = C1 + C2 +…+ Cn .
Halliday, Resnick and Krane, 5th Edition, Chap. 30, MC 9:
The capacitors have identical capacitance C. What is the
equivalent capacitance Ceff of each of these combinations?
A
B
C
D
Halliday, Resnick and Krane, 5th Edition, Chap. 30, MC 9:
The capacitors have identical capacitance C. What is the
equivalent capacitance Ceff of each of these combinations?
A
C
Ceff = 3C
Ceff = 3C/2
Ceff = 2C/3
B
D
Ceff = C/3
Halliday, Resnick and Krane, 5th Edition, Chap. 30, Prob. 9:
Find the charge on each capacitor (a) with the switch open and
(b) with the switch closed.
C1 = 1 μF
C2 = 2 μF
C3 = 3 μF
C4 = 4 μF
ΔV =12 V
Answer (a): We have C1ΔV1 = q1 and C3ΔV3 = q1 since those
two capacitors are equally charged. Now ΔV1 + ΔV3 = 12 V
and so q1/C1 + q1/C3 = 12 V and we solve to get q1 = q3 = 9 μC.
In just the same way we obtain q2 = q4 = 16 μC.
C1 = 1 μF
C2 = 2 μF
C3 = 3 μF
C4 = 4 μF
ΔV =12 V
Answer (b): We have C1ΔV12 = q1 and C2ΔV12 = q2 and in the
same way C3ΔV34 = q3 and C4ΔV34 = q4. Two more equations:
ΔV12 + ΔV34 = 12 V and q1 + q2 = q3 + q4. We solve these six
equations to obtain ΔV12 = 8.4 V and ΔV34 = 3.6 V, and charges
q1 = 8.4 μC, q2 = 16.8 μC, q3 =10.8 μC and q4 = 14.4 μC.
C1 = 1 μF
C2 = 2 μF
C3 = 3 μF
C4 = 4 μF
ΔV =12 V