# The physics of charged objects

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```					ELECTROSTATICS
Outline
• Electric Force, Electric fields

• Electric Flux and Gaub law

• Electric potential

• Capacitors and dielectric (Electric
storage)
The physics of charged objects
• Study of electricity aims to understand
the interaction between different
charged objects.

+               -
The physics of charged objects
• Study of electricity aims to understand
the interaction between different
charged objects.

+               +

-               -
Structure of Matter
• Fundamental building blocks of the
matter are atoms.
-
-
-
-
+ +
+
+ +
-         + +    -

-
Structure of Matter
• Neutral atom – electron = Positive ion

-                               -

1 electron charge  -1.602  10 C          -19
-

+ +
+
+ +
-          + +    -

-
Structure of Matter
• Fundamental building blocks of the
matter are atoms.
-
-
-
-
+ +
+
+ +
-         + +    -

-
Structure of Matter
• Neutral atom + electron = negative ion.

-
-
-
-
+ +
+
+ +
-         + +    -

-

-
ELECTRICALLY CHARGING
OBJECTS

+   -   +   -   +   -
-   +   -   +   -   +
+   -   +   -   +   -
ELECTRICALLY CHARGING
OBJECTS

+   -   +   -   +
-
-   +   -   +   -   +
+   -   +   -   +   -
ELECTRICALLY CHARGING
OBJECTS

+   -   +   -   +   -
-
-   +   -   +   -   +
+   -   +   -   +   -
Charging by Induction
• In metals outer atomic electrons are
not bound to any atoms (electron see).
+
+           +
+               +

+
+
+       +               +
Charging by Induction
• In metals outer atomic electrons are
not bound to any atoms (electron see).
+
+           +
+

-
+
+
+
+       +               +

                        
Charging by Induction
Electric Polarization
• Same atoms have weakly bound
electrons.

-
-
+ +         -
+
+ +
+ +            -
-     -
-
Electric Polarization
• Same atoms have weakly bound
electrons.

-           -
+ +             -       -
+


+ +
+ +
-
-
-


+
Electric Polarization
The Electric Force
Coulomb’s Law
• Quantifies the electric force between
two charges.


Fab  Qa  Qb
Coulomb’s Law
• Quantifies the electric force between
two charges.

       1
Fab  2
rab
Coulomb’s Law
• Quantifies the electric force between
two charges.
       1 QaQb           QaQb
Fab            rab  k 2 rab
ˆ             ˆ
40 rab
2
rab
where
 0  8.854 1012 C 2 / Nm
and
k  8.988 10 Nm / C
9    2    2
Electric Force Field
• Gravitational force field:
Electric Force Field

+Q
Electric Force Field

+Q
+q
Electric Force Field
• Definition of Electric field:


        FQq
E
q
Electric Force Field
• Definition of Electric field:

 kQ
E  2 rQ q
ˆ
rQq
Electric Force Field

+

        r5   
r1            r4        +
+

                       r3
r2
+                           +
Electric Force Field
• The electric field due to a number of source
charges is given by the expression
 N 
E   Ei (ri )
i1
N
qi
 k  2 ri ˆ
i1 ri
Electric Force Field
Electric Force Field
(Linear distribution of charge)
dq
Linear charge density     
dL

r
dq
dE  ?

dE  k
dL     r   2
Electric Force Field
(Linear distribution of charge)
dq
Linear charge density       
dL

dE  ?
dL
dE  k 2
dL  r
Electric Force Field
(Linear distribution of charge)
dq
Linear charge density     
dL

dE  ?
dL
E  k 2
dL  r
Electric Force Field
(Surface distribution of charge)
dq
Surface charge distribution     
da

dE  ?
r          dq
dE  k 2
r
da
Electric Force Field
(Surface distribution of charge)
dq
Surface charge distribution     
da

dE  ?
r         da
dE  k 2
r    da
Electric Force Field
(Surface distribution of charge)
dq
Surface charge distribution     
da

dE  ?
da
r   E k 
surface
da
r 2
Electric Dipole
• Electric dipole consists of a pair of point
charges with equal size but opposite sign
separated by a distance d.

d
+              -

p  q d
Electric Dipole
• Electric dipole consists of a pair of point
charges with equal size but opposite sign
separated by a distance d.

d
+              -
Electric Dipole
• Electric dipole consists of a pair of point
charges with equal size but opposite sign
separated by a distance d.

d
+              -


p
Electric Dipole
• Electric dipole consists of a pair of point
charges with equal size but opposite sign
separated by a distance d.

d
+              -

p  q d
Electric Dipole
• Water molecules are
electric dipoles

+       +

pwater               -
Exercise 1
A point charge q = -8.0 nC is located at the
origin. Find the electric field vector at the point
x = 1.2 m, y = -1.6 m

-

r  2.0m      1.6 m

1.2 m

ˆE j
E  Exi    ˆ
y

-

r  2.0m   1.6 m

1.2 m

ˆ  sin j)
E  E(cos i        ˆ

-


r  2.0m   1.6 m

1.2 m

ˆ  14 N / Cj
E  11 N / Ci           ˆ

-


r  2.0m   1.6 m

1.2 m
Exercise 2
An electric dipole consists of a positive
charge q and negative charge –q
separated by a distance 2a, as shown in
the figure below. Find the electric field due
to these charges along the axis at the
point P, which is the distance y from the
origin. Assume that y>>a.
r           r

q                   q
a   a
Vector Flux
Vector Flux
Vector Flux
• Definition of flux:

  
  v A
Electric Flux
Electric Flux

    
E   E  dA
Gaub’s Law

     Q
Surface E  dA  Enclosed

Enclosed
0
Gaub’s Law

r
+

    
dE  E  dA  E dA
Gaub’s Law

r
+

E   E dA
Gaub’s Law

r
+

kQ
E  2  dA
r
Gaub’s Law

r
+

Q              Q
E            4r 
2

4 0r 2
0
Exercise 3
dE
Solution
Coulomb’s Law

kdq kdA
dE  2 
R    R 2
Solution
Infinitesimal area of disk

dA  2rdr
Solution
Infinitesimal area of disk

k2rdr
dE 
R 2
Solution
Y-component of E-field element

k2rdr
dEy  dE cos          cos
R 2
Solution

L
cos 
R
Solution
Y-component of E-field element

k2rdr  L 
dEy           
R 2
R
Solution
Y-component of E-field element

1
k
4 0
Solution
Y-component of E-field element

  rdr 
dEy     L 2 2 3/ 2 
20  (r  L ) 
Solution
Y-component of E-field element

        rdr 
Ey      L  2 2 3/ 2 
2 0  0 (r  L ) 

rdr          1
Using the identity            (r2  L2 )3/ 2  L
0


Ey 
2 0
Two Oppositely charged Parallel
Plates (Capacitor)
Two Oppositely charged Parallel
Plates (Capacitor)

E?
Exercise 4
Gauss
     Q       L
 E  dA      
0 
0

Area of a cylinder  2rL
L
E 2rL 
0

E 
2r 0
Electric Potential

+Q
+q
Electric Potential

+Q
Electric Potential

Electric potential at a point P due source in its neighborehood

Work on Qtest (  P)
Qtest
Electric Potential

Electric potential at a point P due source in its neighborehood

P 
1           
        F  dr
Qtest 

P 

  E  dr



P
 kQSource ˆ 
          r   dr

r 2

Electric Potential

kQSource QSource
V(r)          
r       4 0r

Unit :              1 V  1J/C
Electric Potential
Electric potential at position P due to a system
of N source charges is given by:

1              N
Qi
VP 
4 0
r
i1  i
Electric Potential
• Potential difference:
Electric Potential
• Potential difference:
Electric Potential

1 1
V  V (rb )  V (ra )  kQsource   
r r 
 b a
rb 

   E  dr
ra
Two Oppositely charged Parallel
Plates (Capacitor)

E  constant

V
Two Oppositely charged Parallel
Plates (Capacitor)
rb     
V        E  dr
ra
rb

 E  dr
ra

 E(ra  r )
b

 Ed
Capacitors and di-electrics
• Capacitors store electric potential energy

Battery

V  V
Charge stored on each plate
Capacitanc e 
Voltage across plates
Q
C
V

1C  1 C / V  1 farad (F)
Capacitors and di-electrics

E
E
Capacitors and di-electrics

Eresultant  E  E


0
Q

 0A
Capacitors and di-electrics
• We can therefore express the voltage
across the capacitor plates as follows:

 Q 
V  Ed      d
  A
 0 

• Hence         Q     Q        A
C          0
V  Qd        d
  A
 0 
Exercise 3
A parallel-plate capacitor has an area of A = 2
cm2 and a plate separation of d =1mm.

(a) Find its capacitance. (answer = 1.77pF)

(b) If the plate separation of this capacitor is
increased to 3 mm, find the capacitance.
Capacitors and di-electrics
• Capacitors in Parallel
Capacitors and di-electrics
• Di-electric material inside a parallel Electric field.
Capacitors and di-electrics
• Di-electric material between parallel capacitor
plates.
Capacitors and di-electrics
• Di-electric material
between parallel
capacitor plates.

Cdielectric  KCvacuum
Capacitors and di-electrics
• Di-electric material between parallel
capacitor plates.

Charge remains constant
Q  CdielectricVdielectric  CvacuumVvacuum
Capacitors and di-electrics
• Di-electric material
between parallel
capacitor plates.

Vdi-electric  KVvacuum
Capacitors and di-electrics
• Di-electric material
between parallel
capacitor plates.

Since V  Ed
 Edi-electric  KEvacuum

```
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