ELECTROSTATICS I

1. Like charges repel, unlike charges attract. But by how much? Coulomb's Law says that
this depends both upon the strength of the two charges and the distance between them.
In mathematical ter 1 2 1 2
2 2 ms, which can be converted into an equality, .
The constant of proportionality will take different values depending upon the units we
choose. In the MKS system, charge is meas
Fq q Fk q q
r r
∝ =
0
12 2 2 9 2 2
0
ured in Coulombs (C) and 1 with
4
8.85 10 / and hence 8.99 10 / .
2. The situation is quite similar to that of gravity, except that electric charges and not masses
are the sou
k
C Nm k Nm C
πε
ε −
=
= × = ×
1 2
12 2 12
0 12
12 1 2
12 21 2 21
12 0 12
rce of force. In vector form, 1 ˆ is the force exerted by 2 on 1,
4
where the unit vector is ˆ . On the other hand, 1 ˆ is the force exerted
4
by 1 on
F q qr
r
rr F qq r
r r
πε
πε
=
= =
􀁇
􀁇 􀁇
12 21
1 12 13 14
2. By Newton's Third Law, . For many charges, the force on charge 1 is
given by,
3. Charge is quantized. This means that charge comes in certain units only. So th
F F
F F F F
= −
= + + + ⋅ ⋅ ⋅ ⋅
􀁇 􀁇
􀁇 􀁇 􀁇 􀁇
19
e size of a
charge can only be 0, , 2 , 3 , where 1.602 10 is the value of the
charge present on a proton. By definition we call the charge on a proton positive. This
makes the c
±e± e± e⋅ ⋅ ⋅ ⋅ e= × − C
harge on an electron negative.
4. Charge is conserved. This means that charge is never created or destroyed. Equivalently,
in any possible situation, the total charge at an earlier time is equal to the charge at a
later time. For example, in any of the reactions below the initial charge = final charge:
(electron and positron annihilate into neutral photons)
e− +e+ →γ +γ
0
2 2 3
(neutral pion annihilates into neutral photons)
(two deuterons turn into tritium and proton)
5. this a quantity that has a defin
H H H p
π →γ +γ
+ → +
Field : ite value at any point in space and at any time. The
simplest example is that of a scalar field, which is a for any value of , , , .
Examples: temperature inside a room ( , ,
single number x y z t
T x y z, ) , density in a blowing wind ( , , , ),
There are also which comprise of three numbers at each value of , , , .
Examples: the velocity of wind, the pressure inside a fl
t xy z t
vector fields, x y z t
ρ ⋅ ⋅ ⋅
uid, or even a sugarcane field. In
1 2 3 every case, there are 3 numbers: ( , , , ) { ( , , , ), ( , , , ), ( , , , ) }.
6. The electric field is also an example of a vector field, and will be the most important for
our purpose.
V x y z t = V x y z t V x y z t V x y z t
􀁇
0
0
It is defined as the force on a unit charge. Or, since we don't want the charge
to disturb the field it is placed in, we should properly define it as the force on a "test"
charge q, E F . H
q
≡ 0
0
2
0
2
0 0
ere is very very small. The electric field due to a point charge can
be calculated by considering two charges. The force between them is 1 and so
4
1 . A way to visualiz
4
q
F qq
r
E F q
q r
πε
πε
=
= = e E fields is to think of lines starting on positive charges
and ending on negative charges. The number of lines leaving/entering gives the amount of
charge.
7. Typical values for the maginitud
11
5
e of the electric field :
Inside an atom- 10 N/C
Inside TV tube- 10 N/C
E
2
-2
In atmosphere- 10 N/C
Inside a wire- 10 N/C
8. Measuring charge. One way to do this is to balance the gravitational force pulling a
charged particle with mass m with the force exerted on it by a known electric field
(see below). For equilibrium, the two forces must be equal and so . The
unknown
mg = qE
charge q can then be found from qmg .
E
=
1 2 3
9. Given several charges, one can find the total electric field at any point as the sum of the
fields produced by the charges at that point individually, or
i
i
i i
E E E E
q E E k
r
= + + + ⋅ ⋅ ⋅ ⋅ ⋅
=Σ =
􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 ( ) 2ˆ 1,2,3, . Here ˆ is the unit vector pointing from the charge to
the point of observation.
i i
i
Σ r i= ⋅ ⋅ ⋅ r
E 􀁇
eE
􀁇
mg􀁇
10. Let us apply the principle we have just learned to a system of two charges and -
which are separated by a distance (see diagram). Then, . Just to make
things easier (not
q q
d EE E + −
+
= +
􀁇 􀁇 􀁇
2 2 ( )2
0 0
necessary; one can do it for any point) I have taken a point that lies
on the x-axis. The magnitudes of the electric field due to the two charges are equal;
1 1 . The
4 4 / 2
E E q q
+ − πε r πε x d = = =
+
2 ( )2
vertical components cancel out, and the net
electric field is directed downwards with magnitude, cos cos 2 cos .
From the diagram below you can see that cos / 2 . Substituting
/ 2
E E E E
d
x d
θ θ θ
θ
+ − + = + =
=
+
( ) ( ) ( ) 02 22 2 02 23/ 2
this, we
find: 21 /2 1 .
4 / 2 / 2 4 / 2
E q d qd
πε x d x d πε x d
= =
+ + ⎡⎣ + ⎤⎦
11. The result above is so important that we need to discuss it further. In particular, what
happens if we are very far away from the dipole, meaning ? Let us first define
the
x d
dipole
>>
( )
2 3 / 2
3 2 3 / 2 3 3
0 0 0
as the product of the charge the separation between them .
Then, 1 1 1 1 1 . In the above,
4 1 /2 4 2 4
/ 2 has been neglected in
moment p qd
E p p d p
x d x x x x
d x
πε πε πε
−
× =
= = ⎡⎢ +⎛⎜ ⎞⎟ ⎤⎥ =
⎣⎡ + ⎦⎤ ⎢⎣ ⎝ ⎠ ⎥⎦
3
0
comparison to 1. So finally, we have found that for ,
1 .
4
12. It is easy to find the torque experienced by an electric dipole that is placed in a uniform
electric field: The magni
x d
E p
πε x
>>
=
( )
tude is sin sin sin , and the direction is
2 2
perpendicular and into the plane. Here is the angle between the dipole and the electric
field. So sin sin .
Fd Fd Fd
qE d pE
τ θ θ θ
θ
τ θ θ
= + =
= =
θ
θ θ θ
r
r
d
x
E−
􀁇
E
􀁇
+
−
E+
􀁇
dE cosθ
dE
􀁇
θ
θ
R
x
y
z
λ ds
P