Ben Gurion University of the Negev www.bgu.ac.il/atomchip , www.bgu.ac.il/nanocenter
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
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
Sources: Halliday, Resnick and Krane, 5 th Edition, Chaps. 28, 30.

With 100,000 V on her body, why is this girl smiling???

E from V
Let U ( r
2
) be the potential energy of a charge q at the point r
2
.
We found that it is minus the work done by the electric force
F q in bringing the charge to r
2
:
U ( r
2
)

U ( r
1
)
  r
2

F q
( r )
 d r .
r
1
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 :
V ( r
2
)

V ( r
1
)
  r
2

E ( r )
 d r .
r
1

E from V
Let V ( r
2
) be the electric potential at the point r
2
. We have just obtained it from the electric field E :
V ( r
2
)

V ( r
1
)
  r
2

E ( r )
 d r .
r
1
Now let’s see how to obtain can write r
2
= r
1
E from V . For r
2 close to r
1
+ Δ r and expand V ( r
2
) in a Taylor series: we
V ( r
2
)

V ( r
1
  r )

V ( r
1
)


V
 x x
1
 x


V
 y y
1
 y


V
 z z
1
 z

...
;

E from V
We can write this expansion more compactly using the “del”
(gradient) operator:
 



 x
,

 y
,

 z

 , so

1
)




V
 x
,

V
 y
,

V
 z

 , with all the derivatives evaluated at the point r
1
. We also write

V ( r
1
)
  r


V
 x
 x


V
 y
 y


V
 z
 z .

E from V
So now we can write
V ( r
1
  r )

V ( r
1
)


V
 x
 x


V
 y
 y


V
 z
 z

...

V ( r
1
)
 
V ( r
1
)
  r

...
and therefore
V ( r
2
)

V ( r
1
)

V ( r
1
  r )

V

V
( r
1
)

( r
1
)
 r


 r
2

E ( r )
 d r r
1

E ( r
1
)

E ( r
1
)
  r
  r .

E from V
The equation

V ( r
1
)
  r
 
E ( r
1
)
  r is a scalar equation, but since we can vary each component of
Δ r independently, it actually yields three equations:


V ( r
1
)
 x
 
E x
( r
1
) ,


V ( r
1
)
 y
 
E y
( r
1
) ,


V ( r
1
)
 z
 
E z
( r
1
) ,

E from V
The equation

V ( r
1
)
  r
 
E ( r
1
)
  r is a scalar equation, but since we can vary each component of
Δ r independently, it actually yields three equations, i.e.

 x

 y
V ( r
1
V ( r
1
)
)

 z
V ( r
1
)
 
E x
( r
1
) ,
 
E y
( r
1
) ,
 
E z
( r
1
) ,

E from V which reduce to a single vector equation:
E ( r )
 
V ( r ) .

 x

 y
V ( r
1
V ( r
1
)
)

 z
V ( r
1
)
 
E x
( r
1
) ,
 
E y
( r
1
) ,
 
E z
( r
1
) ,

E from V
Example 1 (Halliday, Resnick and Krane, 5 th Edition, Chap. 28,
Exercise 34): Rutherford discovered, 99 years ago, that an atom has a positive nucleus with a radius about 10 5 times smaller than the radius R of the atom. He modeled the electric potential inside the atom ( r < R ) as follows:
V ( r )

4
Ze

0


1 r

3
2 R
 r
2
2 R
2
, where Z is the atomic number (number of protons). What is the corresponding electric field?

E from V
Answer: E ( r )
 
V ( r )
 
4
Ze

0

1 r

3
2 R
 r
2
2 R
2


4
Ze

0


1 r
2
 r
R
2


 r

4
Ze

0


1 r
2
 r
R
2

 r
ˆ
, since
 r

 r
 x

 r
 y

 r
 z
 x
 r y
 z 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 d /2 z r
–
( x , y , z )
V ( x , y , z )
 q
4

0


1 r 

1 r 


, r
+ whe re r 
 x
2  y
2 
( z
 d
2
)
2
.
– d /2

E from V
E x
( r )
 

V
 x
 qx
4

0



 r 
1 r 
1
   
3




,
E y
( r )
 

V
 y
 qy
4

0



 r 
1 r 
1
   
3




, d /2 z r
–
( x , y , z )
E z
( r )
 

V
 z
 q
8

0




2 z r 

 
3 d

2 z r 

 
3 d




, r
+
whe re r 
 x
2  y
2 
( z
 d
2
)
2
.
– d /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 r 
 x
2  y
2  
 d
2


2
 x
2  y
2  z
2  zd
 d
2
4 d /2 z
 r




1
 zd r
2
 d
2
4 r
2


 1/2
, where r
 x
2  y
2  z
2
.
r
–
( x , y , z )
– d /2 r
+

E from V
For d << r
–
, r
+ we have r 
 r

1
 

1/2
, where
   zd r
2
 d
2
4 r
2
, so
1
( r  )
3

1 r
3

1
 
 
3/2 
1 r
3

 
3
2




1 r
3




1 
3 zd
2 r
2

3 d
2
8 r
2




.
d /2 z
– d /2 r
– r
+
( x , y , z )

E from V
So r 
1
3
 r 
1
   
3

1 r
3




1

3 zd
2 r
2

3 d
2
8 r
2




1 r
3




1

3 zd
2 r
2

3 d
2
8 r
2




3 zd
; and now r
5
E x

4
3 xzqd

0 r
5
, E y

4
3 yzqd

0 r
5
, E z

( 3 z
2
4

 r
2
0 r
5
) qd
.
d /2 z r
–
( x , y , z )
– d /2 r
+

E from V
E x

4
3 xzqd

0 r
5
, E y

4
3 yzqd

0 r
5
, E z

( 3 z
2
4

 r
2
0 r
5
) qd
.
We can check that these results coincide with the results we obtained for the special cases x = 0 = y and z = 0:
E z

2 qd

0 z
3
, E z
 
4 qd

0 r
3
.
d /2 z r
–
( x , y , z )
– d /2 r
+

E from V
E x

4
3 xzqd

0 r
5
, E y

4
3 yzqd

0 r
5
, E z

( 3 z
2
4

 r
2
0 r
5
) qd
.
This may look complicated but it is easier than calculating E x
,
E y and E z directly! d /2 z
– d /2 r
– r
+
( x , y , z )

Equipotential surfaces
We have already seen equipotential surfaces in pictures of lines of force: and in connection with potential energy:
F ( r ) r

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

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
V
We can also visualize the topography of the electric potential from the side (left) and from above (right) with equipotentials as horizontal 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
9 V
8 V
7 V
6 V
A
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 /ε
0
, 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 /ε
0
, 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 and the other on a thin 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 2 EA = Ф
E
= σA /ε
0
, so E = σ /2ε
0
. But for a conductor, there is flux only through the top of the can, so Gauss
’ s law gives
EA = Ф
E
=
σA
/ε
0 and E =
σ
/ε
0
.
Conclusion: For the same surface charge σ , the electric field on a conductor is twice the electric field on a thin charged sheet.
“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 /ε
0
, 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· d r 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/ r 2 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
· d r 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 Δ V across the battery terminals: q = C Δ V. 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 Δ V across the battery terminals: q = C Δ V. (Note that q and Δ V 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 Δ V across the battery terminals: q = C Δ V. (Note that q and Δ V 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
/ε
0
Δ
V = Ed = qd /ε
0 where
σ
= q /
A . By definition, q = C
Δ
V , hence C = ε
A
0 and
A / 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
E ( r )

1
2

0 q
Lr
, a

r

b ,

V

2
 q
0
L
 b a dr r

2
 q
0
L ln ( b/a ) ,
C

2
 ln b
0
L
 ln a
.
a b

Capacitors and capacitance
Example 3: Spherical capacitor
The diagram is unchanged, only E ( r ) is different:
E ( r )

1
4

0 q r
2
, a

r

b ,

V
 q
4

0
 b a dr r
2
 q
4

0

 a
C

4
 b

0 ab a
.
1 b


, 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 Δ V , we must have q
1
= C
1
Δ
V
1 and q
2
= C
2
Δ
V
2 where Δ V
1
+ Δ V
2
= Δ V . But the charge on one capacitor comes from the other, hence q
1
= q
2
= 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 Δ V , we must have q = C
1
Δ
V
1
Δ
V
1
+ Δ V
2 and
= Δ V q = C
2
Δ
V
2 where
. If the effective capacitance is C eff
, then we have q
C eff
 
V
 
V
1
 
V
2
 q
C
1
 q
C
2
,
1 hence
C eff

1
C
1

1
C
2
.

Capacitors in series and in parallel
For n capacitors in series, the generalized rule is
1
C eff

1
C
1

1
C
2

...

1
C n
.
C
1
C
2
C n

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 Δ V , the charge on the capacitors is q
1 and q
2
= C
2
= C
1
Δ
V
Δ V . If the effective capacitance is C eff
, then we have
C
1
Δ
V + C
2
Δ
V = q
1
+ q
2
= C eff
Δ
V, thus capacitances in parallel add :
C
1
+ C
2
= C eff
.

Capacitors in series and in parallel
For n capacitors in parallel, the generalized rule is
C eff
= C
1
+ C
2
+…+ C n
.
C
1
C
2
C n

Halliday, Resnick and Krane, 5 th Edition, Chap. 30, MC 9:
The capacitors have identical capacitance C . What is the equivalent capacitance C eff of each of these combinations?
A B
C D

Halliday, Resnick and Krane, 5 th Edition, Chap. 30, MC 9:
The capacitors have identical capacitance C . What is the equivalent capacitance C eff of each of these combinations?
C eff
= 3 C
C eff
= 2 C /3
A B
C
C eff
= 3 C /2
D
C eff
= C /3

Halliday, Resnick and Krane, 5 th Edition, Chap. 30, Prob. 9:
Find the charge on each capacitor (a) with the switch open and
(b) with the switch closed.
Δ V = 12 V
C
1
= 1 μF
C
3
= 3 μF
C
2
= 2 μF
C
4
= 4 μF

Answer (a): We have C
1
Δ V
1
= q
1 and C
3
Δ V
3 two capacitors are equally charged. Now Δ V
1
= q
1
+ Δ since those
V
3
= 12 V and so q
1
/ C
1
+ q
1
/ C
3
= 12 V and we solve to get
In just the same way we obtain q
2
= q
4
= 16 μC.
q
1
= q
3
= 9 μC.
Δ V = 12 V
C
1
= 1 μF
C
3
= 3 μF
C
2
= 2 μF
C
4
= 4 μF

Answer (b): We have C
1
Δ V
12 same way C
3
Δ
V
34
Δ
V
12
+ Δ V
34
= q
3 and
= 12 V and q
1
+
C q
4
Δ
V
34
2
= q
1
= q
3 and C
2
Δ V
12
=
+ q q
4
4
= q
2 and in the
. Two more equations:
. We solve these six equations to obtain Δ V
12 q
1
= 8.4 μC, q
2
= 8.4 V and Δ
= 16.8 μC, q
3
V
34
= 3.6 V, and charges
=10.8 μC and q
4
= 14.4 μC.
Δ V = 12 V
C
1
= 1 μF
C
3
= 3 μF
C
C
2
4
= 2 μF
= 4 μF