DO PHYSICS ONLINE
ELECTRICAL TERMINOLOGY REVIEW
To understand the concepts in the Module From Ideas to Implementation it is essential that
you have a good working knowledge of basic electrical and magnetic language, concepts and
mathematical representation. This unit gives a brief summary of the most important ideas.
ATOM nucleus (protons + neutrons) & electron- cloud
CHARGE Q, q [C coulomb]
charge of electron = -1.610-16 C charge of proton = + 1.610-16 C
1 C = 6.261018 electrons
transfer of electrons  positive and negative charged objects
rubber rod -
glass rod +
e-
e-
wool +
silk -
FORCE BETWEEN CHARGES
charged
rod
+
neutral
DO PHYSICS ONLINE
+
same charges
repel
+
-
opposite charges
attract
neutral pieces
of paper
any charged object
will attract a neutral
object
1
F
Coulomb’s Law
1
q1 q2
4  o r 2
q2
q1
vacuum
r
q1 q2 > 0
F > 0 repulsive force
q1 q2 < 0
F < 0 attractive force
ELECTRIC FIELD E [N.C-1 or V.m-1]
E
region surrounding charge distribution
+q
F = E q or F  q E
+
Electric field
POSTIVE charge
F
+
+
+
+
+
-
-
-
-
-
-
Electric field
NEGATIVE charge
Uniform electric field
E = V/d
DO PHYSICS ONLINE
2
POTENTIAL DIFFERENCE, POTENTIAL, VOLTAGE V [V VOLT]
VOLTAGE DROP, POTENTIAL DROP, EMF
++++++
+6V
-3V
+6V
+++
+3V
+ 15 V
0V
-2 V
-+6V
-14 V
- -------
-8V
-2V
------------
-12 V
potential differences
potentials
- measure of charge imbalance
- A charge q of mass m in a electric field E, accelerated by a voltage V increases its kinetic
energy EK by
q V = ½ m v2
To a good approximation a uniform electric field is produced when two parallel plates with a
small separation between them are connected to a battery. Charge is stored on each plate, one
+Q and the other –Q. The electric field is equal to the gradient of the potential
E
V
x
and for a uniform electric field
E
V
d
where V is the potential difference between the plates and d is the separation distance
+ + + + + + + + + +
V
-
Do problems:
d
E
p1.93
DO PHYSICS ONLINE
-
- -
-
-
-
-
-
-
p1.28
3
ELECTRIC CURRENT
I i [A ampere
A = C.s-1]
The flow of charge gives rise to an electric current. The current is defined by the equation
q
I
t
where q is the amount of charge that passes a cross-section in the time interval t.
I = nqvA
where
n = number of charge carriers / volume [m-3]
q = charge on particle [C]
v = drift velocity of charged particles [m.s-1]
A = cross sectional area [m2]
Direction of current I - direction in which positive charges would move
DC – current in the one direction or polarity (+ / -) of voltage does not change
AC – direction of current oscillates or polarity (+ / -) of voltage oscillates
DO PHYSICS ONLINE
4
RESISTANCE
R
[
ohm]
The opposition to the movement of charges (current) due to energy being dissipated as
thermal energy is known as resistance.
I
V
R
R
A
L
 resistivity (property of material [.m]
A cross sectional area [m2]
L length [m]
Resistors in series
Resistors in parallel
R
R  R1  R2 
R1
R2
1
1
1
1


R1 R2 R3
...

R1
R
R2
...

Adding extra resistors increases the
equivalent resistance
POWER
ENERGY
TIME
P
W
t
R
Adding extra resistances decreases the
equivalent resistance
[W watts]
[J joule]
[s
second]
The rate at which electrical energy is consumed is known as the power P (e.g. 100 W light
globe).
W
V2
V I 
 I 2R
t
R
where W is the thermal energy dissipated in a time interval t.
P
Energy source: battery or power supply
R
I
emf 
energy supplied by battery W =  q
power supplied by battery
DO PHYSICS ONLINE
Energy dissipated by a resistance
P = W/t =  I
I
voltage drop V
energy dissipated W = V I t
power dissipated P = V I
5
MAGNETIC FIELD STRENGTH (MAGNETIC FLUX DENSITY)
B
[T TESLA]
North
geographic pole
N magnetic
pole
S magnetic
pole
South
geographic pole
I
I
B
right hand
screw rule
B
I
solenoid
current
out
current
in
uniform magnetic field
DO PHYSICS ONLINE
6
FORCE ON A CURRENT CARRYING CONDUCTORS IN A MAGNETIC
FIELD
F  B I L sin 
FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD
F  B q v sin 
+q
+I
F
out of page
B
right hand palm rule
F
F
force on current carrying conductor [N]
force on charge particle
magnitude only
B
strength of uniform magnetic field [T tesla]
magnetic flux density
magnitude only
I
q
convectional current through conductor [A ampere]
charge of particle [C coulomb]
magnitude only
magnitude only
L
length of conductor in magnetic field [m]
magnitude only
v
velocity of charged particle [m.s-1]
magnitude only

angle between magnetic field lines and current direction
 = 0  F = 0 current direction & magnetic field
direction parallel
o
 = 90  F = Fmax current direction perpendicular to
magnetic field direction
magnitude only
DO PHYSICS ONLINE
7
B
B
I
Do problems:
p1.34
DO PHYSICS ONLINE
p1.79
8