Physics II - Spring 2021

Chapter 24: The Electric Potential

The following subjects will be discussed

• Definitions: Electric Potential Energy .

• Potential-field Relations.
• Potential in the space of an isolated point charge.
• Potential in the space of many point charges.
• Potential due to continuous charge distribution.
• Equipotential lines/surfaces.
• Potential on Conductors.
 Example

Two tiny particles of mass 𝑚! =5 g and 𝑚" =10 g both with the same charge of 𝑞=5
µC, are connected with an insulating string of length 𝑑=1 m.

1) Calculate the potential energy stored in this system.

𝑞! 1
𝑈=𝑞𝑉= = 0.225 𝐽
4𝜋𝜀" 𝑑

2) Find the acceleration of the charges at the instant the string is cut.
𝐹 𝑞! 1 1 𝐹 𝑞! 1 1
𝑎# = = ! = 45 𝑚/𝑠 ! 𝑎$ = = ! = 22.5 𝑚/𝑠 !
𝑚 4𝜋𝜀" 𝑑 𝑚# 𝑚 4𝜋𝜀" 𝑑 𝑚$

3) After a long time since the string was cut, find the the particles velocities.
1 1
Conservation of energy gives 𝑚# 𝑣#! + 𝑚$ 𝑣$! = 𝑈 = 0.225 𝐽
2 2
1
Also 𝑣$ = 𝑣
2 # 𝑣# = 7.75 𝑚/𝑠

𝑣$ = 3.875 𝑚/𝑠
 Example

A very thin finite line with uniform charge density 𝜆 (C/m) and length L, placed along
the x-axis, as in the figure.
Find an expression for the electric potential at any point along the negative x-axis.

𝑑𝑞
𝑑𝑉% = 𝑑𝑞 = 𝜆 𝑑𝑙 = 𝜆 𝑑𝑥
4𝜋𝜀"𝑟 𝑥 𝑑𝑙 = 𝑑𝑥
&
𝜆 1
𝑉% = = 𝑑𝑥
4𝜋𝜀" 𝑑 + 𝑥 𝑟 =𝑑+𝑥
"

𝜆 𝑑+𝐿
𝑉% = ln
4𝜋𝜀" 𝑑
 Equipotential Lines/Surfaces

For a given charged structure, if we connect all points that are at the same potential,
we get what is called equipotential lines (2D analysis) or surfaces (3D analysis)

Here are some examples

A point charge

At any given 𝑟, if we connect all the

points with the same potential, we get a
circle (for 2D analysis) or a spherical
surface (for 3D analysis).
Equipotential
Remember that the potential for a point Line/Surface
charge is given by
Field Lines
𝑞 1
𝑉! = The equipotential lines are
4𝜋𝜀# 𝑟
always perpendicular to the
field lines. Why?
 Equipotential Lines/Surfaces

A uniform field (such as inside a parallel plate capacitor)

At any given 𝑥, the path between A and

B is perpendicular to the field lines. So,
the dot product is zero.
"
A
𝑉" − 𝑉! = − , 𝐸 . 𝑑𝑙 = 0
!
B
So, A and B are at the same potential. This
is true at any location 𝑥.

Notice that in 2D analysis, we get vertical 𝑥

equipotential lines. In 3D analysis, we get
equipotential surfaces which are rectangular The equipotential lines are
planes. always perpendicular to the
field lines. Why?
 Equipotential Lines/Surfaces

An Electric Dipole

The equipotential lines can be

drawn by tracing the field lines
and making sure the two are
normal to each other at any
Equipotential
point.
Line/Surface

The equipotential lines are

always perpendicular to the
field lines. Why? Field Lines
 Potential on and Around Conductors

Uncharged Conductors
If an initially uncharged conductor is placed
within the effect of some electric field, as in
the picture, then

1) The electric field everywhere on the

conductor is ZERO. The external field
will separate the free charges of the
conductor to either side (depending on
the external field direction).
2) That means the potential everywhere on
the conductor is the SAME (not zero).
External Electric Field
3) The field lines are perpendicular to the
surface of the conductor.
 Potential on and Around Conductors

Charged Isolated Conductors

For an isolated conductor of any shape, if
we put some charge on it with no external
field, as in the picture, then
cavity
1) The electric field everywhere on the
conductor is ZERO. The external
charge will all be located on the
surface.
2) The distribution of the electric No External Electric Field
charge on the conductor surface will
not be uniform, in general. Flat parts 4) The situation does not change even
have less charge density than curved if there is a cavity within the
parts (why?). conductor.
3) That means the potential everywhere
on the conductor is the SAME (not
zero).
 Potential on and Around Conductors

Charged Isolated Conductors

An uncharged conductor (right) is brought in the vicinity of a charged conductor (left).
 Next Two Chapters

Based on the subjects we studied regarding

Charges, Electric Field and Potential

Two Circuit Components will be analyzed

(1) The capacitor: Based on stored charges

Chapter 25.

(2) The resistor: Based on transported (moved) charges

Chapter 26.
 Physics II

Chapter 25: Capacitance

The following subjects will be discussed

• Definition of Capacitance.
• Capacitance calculations of several structures.
• Energy stored in capacitors.
• Capacitors in electric circuits.
• Examples.
 Capacitors

+
+ + +
+ +
+ + + E
𝐶
- - -
- -
-
Capacitors
 Capacitors

We can find capacitors in almost every electronic circuit; analog or digital. They serve
many purposes, such as:

Temporary storage of charge/energy.

Timing

Coupling circuits together to block DC/allow AC

Filtering and signal processing

 Definition of Capacitance

The amount of charge that can be stored for every 1 volt of potential difference, is
called Capacitance.
𝑄 Capacitance does not
𝐶= Unit is Farad (F) mean capacity.
𝑉
A capacitor is not a
Depends on (1) shape (2) dimensions (3) filling material. tank!

Example: The parallel-plate capacitor

𝐴
𝐶 = 𝜀$
𝑑
Discharging
Charging
 Parallel-plate Capacitor Example

Consider a parallel-plate capacitor with plates of dimensions 2 cm x 2 cm and

separation of 1 mm. The filling dielectric is air.
Calculate
1) The capacitance of this structure.
! #.#(×#.#(
𝐶 = 𝜀$ = 8.85×10&'( = 3.54 pF
% #.##'

2) The maximum charge it can handle. 𝑄=𝐶𝑉 𝑄+,- = 𝐶 𝑉+,- = 𝐶 𝐸+,- 𝑑

𝑄+,- = 3.54×10&'( ×0.001×3×10. = 0.0106 µC

3) Repeat for paper filling (𝜀! =3.7).

! #.#(×#.#(
𝐶 = 𝜀$ 𝜀/ = 8.85×10&'( ×3.7 = 13.1 pF
% #.##'
From week 3
𝑄+,- = 13.1×10&'( ×0.001×15×10. = 0.20 µC
20 Times more!!
 Capacitance Calculations:
Parallel-plate Capacitor
To find the capacitance of any given structure, the procedure is as follows.
1) Identify the two surface between which the capacitance is to be found.
2) Assume 𝑄0 and 𝑄& on the two surfaces.
3) Find the corresponding electric field 𝐸 .
4) Find the potential difference using 𝑉 = − ∫ 𝐸 . 𝑑𝑙.
5) Use 𝐶 = 𝑄/𝑉. 𝐴
𝑄(
Let’s apply it to the parallel-plate structure with 𝑑
𝐸
𝑑 ≪ other dimensions (the plates can be considered
infinite) 𝑄'
1) Top and bottom.
2) Assume 𝑄0 and 𝑄& on the two surfaces.
1! &1 #
3) The electric field due to bottom plate is 𝐸 = 𝑎⃑ ,
(2"! 3
top plate 𝐸 = (2"!
(−𝑎⃑3 ),
1
total field between the plates 𝐸 = 2 ! 𝑎⃑3 .
"
1%
4) The potential is obtained by integrating 𝐸 from one plate to the other 𝑉 = 2 !.
"
1 2" !
5) Use 𝐶 = =4 %
.
 Capacitance Calculations:
Cylindrical Capacitor
Let’s apply the procedure to the cylindrical structure below.

𝑄(
The electric field in the region 𝑎 ≤ 𝑟 ≤ 𝑏 using
Gauss’ Law. 𝑄'
The charge enclosed = 𝑄
The surface area = 2𝜋𝑟𝐿
𝐿
1
So, the field inside the cylinder 𝐸=
(52"/6
𝑎⃑/

The potential 𝑉 = − ∫ 𝐸 . 𝑑𝑙. Integrate from 𝑎 to 𝑏.

&1 7' 1
𝑉= ∫ 𝑑𝑟 𝑉= ln(𝑏/𝑎)
(52"6 , / (52"6

1 (52 6
"
Now. 𝐶 = 4 . 𝐶 = 89(7/,)
 Capacitance Calculations:
Spherical Capacitor
Let’s apply the procedure to the spherical structure below.

The electric field in the region 𝑎 ≤ 𝑟 ≤ 𝑏 using

Gauss’ Law.
The charge enclosed = 𝑄
𝑄(
The surface area = 4𝜋𝑟 ( 𝑄'

1
So, the field inside the cylinder 𝐸=
=52"/ $
𝑎⃑/

The potential 𝑉 = − ∫ 𝐸 . 𝑑𝑙. Integrate from 𝑎 to 𝑏.

&1 7 ' 1 ' '

𝑉= ∫ 𝑑𝑟 𝑉= −
=52" , / $ =52" , 7

1 ,7
Now. 𝐶 = 4 . 𝐶 = 4𝜋𝜀$
7&,
 Example: A Design Consideration

Consider a cylindrical capacitor with 𝑎, 𝑏 = 2𝑎 and 𝐿.

2𝜋𝜀) 𝐿
𝐶=
ln(𝑏/𝑎)

To increase the capacitance, which is more

efficient 10% change in 𝑎 or 𝐿?

2𝜋𝜀) 𝐿
10% in change in 𝑎 𝐶* =
ln(2/1.1)

2.2 𝜋𝜀) 𝐿
10% in change in 𝐿 𝐶! =
ln(2)
𝐶* 1.16
= >1
𝐶! 1.1

Separation more
effective
 Energy Stored in Charged Capacitors

Starting with 𝑈 = 𝑞𝑉, Can we say that the energy stored in a capacitor 𝑈 = 𝑄𝑉 ?

No!

Because the charge does not appear on the capacitor instantaneously, the correct
equation is 𝑑𝑈 = 𝑉 𝑑𝑞, and we integrate as

+ +𝑞 1 𝑄!
𝑈 = = 𝑉 𝑑𝑞 = = 𝑑𝑞 =
" " 𝑐 2 𝐶

1 𝑄! 1 1
𝑈= 𝑈= 𝑄𝑉 𝑈= 𝐶 𝑉!
2 𝐶 2 2

The energy equations apply for any capacitor structure.

 Capacitors with Dielectrics

For any solid or liquid dielectric 𝜀 = 𝜀/ 𝜀$ , where 𝜀/ > 1.

For example, paper 𝜀/ = 3.7 , water 𝜀/ = 80 , glass 𝜀/ = 5.4.

Why do we use dielectric as capacitor filling between conductors?

𝐴
1) Increase capacitance. 𝐶 = 𝜀$ 𝜀/
𝑑
2) Increase rigidity of the structure.
3) Increase structure electrical strength (breakdown).

4) The energy stored is less. 𝑈 = 𝑈$ /𝜀/ Why?

+ + + +
_ _
Material
+
Polarization
+
- - - -
 Capacitors in Electrical Circuits

Single Series Parallel

𝐶 𝐶* 𝐶! 𝐶, 𝑄*

𝐶*
𝑄 𝑄 𝑄 𝑄
𝑄!
𝑉* 𝑉! 𝑉,
𝐶!
𝑄,

𝐶,
𝑉 𝑉

𝑄 𝑉 = 𝑉* + 𝑉! + 𝑉, 𝑉
𝐶=
𝑉 𝑄 𝑄 = 𝑄* + 𝑄! + 𝑄,
𝐶-)- =
𝑉* + 𝑉! + 𝑉,
𝑄* + 𝑄! + 𝑄,
1 𝐶-)- =
𝐶-)- = 𝑉
1 1 1
+ +
𝐶* 𝐶! 𝐶, 𝐶-)- = 𝐶* + 𝐶! + 𝐶,
 Capacitors in Electrical Circuits

Example 𝐶*
𝑄* = 𝑄 = 6.4 𝜇𝐶
𝐶* + 𝐶!
𝑄*
3.55 µF 𝑉* = = 1.8 𝑉
𝐶*
𝐶
𝐶*
𝑄
𝑄 = 𝐶𝑉 3.55 µF
𝑄 = 22.4 𝜇𝐶
8.95 µF
Remove battery
Connect
𝑉 another 𝐶!
6.3 V capacitor 𝐶!
𝑄! = 𝑄 = 16 𝜇𝐶
𝐶* + 𝐶!
Initial circuit 𝑄!
𝑉! = = 1.8 𝑉
𝐶!
 Capacitors in Electrical Circuits

Example

A Two capacitors in series

𝐴
𝐶>,?@ = 𝜀
(𝑑 − 𝑎)/2
a conductor d
𝐴
𝐶A$A = 𝜀
(𝑑 − 𝑎)

Capacitance
𝐴 increased
𝐶B3BAB,C =𝜀
𝑑
 Capacitors in Electrical Circuits

Example: Find charge and potential difference on each capacitor. 𝑄=𝐶𝑉

𝐶-)- = 12 + 5.3 = 17.3 𝜇𝐹

12 µF 5.3 µF 13.8 µC 2.58 V 44.6 µC

2.58 V 17.3 µF 3.57 µF
30.8 µC
12.5 V 12.5 V
12.5 V
44.6 µC
4.5 µF
4.5 µF
9.92 V 44.6 µC 9.92 V 44.6 µC

1
𝐶-)- = = 3.57 𝜇𝐹
1 1
+
17.3 4.5
 Capacitors in Electrical Circuits

Example: Consider the following capacitor situation.

𝐴 = 115 𝑐𝑚!

𝜀. = 1

𝑉) = 85.5 V 𝑎 = 0.78 𝑐𝑚 Dielectric 𝜀. = 2.61 𝑑 = 1.24 𝑐𝑚

1) Find the initial C, Q, V and E.

2) Find the new C, Q, V and E.
 Capacitors in Electrical Circuits

1) Find the initial C, Q, V and E.

𝐴 = 115 𝑐𝑚!

𝐴 8.85×10(*!×115×10(/ Q
𝐶) = 𝜀) =
𝑑 1.24×10(!
𝜀. = 1
𝐶) = 8.21 pF
𝐸 𝑑 = 1.24 𝑐𝑚

𝑉) = 85.5 V

𝑄) = 𝐶) 𝑉) = 8.21×10(*!×85.5 Q

𝑄) = 702 pC
𝑉) 85.5
𝐸) = =
𝑑 1.24×10(!

𝐸) = 6.9 kV/m
 Capacitors in Electrical Circuits

2) Find the new C, Q, V and E. 𝐴 = 115 𝑐𝑚!

Q
Charge stays the same! 𝑄 = 702 pC
The electric field inside the dielectric 𝜀. = 1 𝐸

𝑎 = 0.78 𝑐𝑚

𝑑 = 1.24 𝑐𝑚
𝐸 6.9×10, Dielectric 𝜀. = 2.61 𝐸"
𝐸0 = = = 2.64 kV/m
𝜀. 2.61
The potential across the capacitor 𝐸
1 Q
𝑉 = − = 𝐸 𝑑𝑙 = 𝐸 𝑑 − 𝑎 + 𝐸0 𝑎 The electric field inside
"
the dielectric will be
𝑉 = 6.9×10,× 1.24 − 0.78 ×10(! + 2.64×10,×0.78×10(! different!

𝑉 = 52.3 V Why decreased?

𝑄 702×10(*!
𝐶= =
𝑉 52.3
Can you guess the
𝐶 = 13.4 pF Why increased? answer here?
 Capacitors in Electrical Circuits

Example: After steady-state, C3 undergoes breakdown (becomes short circuit). Find

the change in charge and potential across C1.
213.3 µC
𝐶* 10 µF
100 V 𝐶-)-
5 µF 𝐶! 21.33 V 100 V 15 µF 𝐶* + 𝐶!
100 V 320 µC
106.7 µC 320 µC
320 µC 320 µC 3.2 µF

𝐶, 𝐶, 4 µF 1
4 µF
𝐶-)- =
1 1
+
𝐶* + 𝐶! 𝐶,
𝐶* 10 µF
After breakdown
100 V 𝐶! 1000 µC
What is the message
100 V
in this example?
5 µF
 Capacitors in Electrical Circuits

Example: After the two capacitors are connected, find the change in charge and
potential across each one.

_
1 µF + _
3 µF 1 µF _ 3 µF
𝐶* 𝐶! 𝐶* 𝐶!
_ 50 µC + + 150 µC
100 µC + 300 µC

100 V 100 V 50 V 50 V

Total charge is
300 µC -100 µC =200 µC.
It will divide based on capacitance.
 Q&A

Send me an email if
you have any
questions.