ELECTROSTATICS II

1. In the last lecture we learned how to calculate the electric field if there are any number
of point charges. But how to calculate this when charges are continuously distributed over
some reg
1 2 3
ion of space? For this, we need to break up the region into little pieces so that
each piece is small enough to be like a point charge. So, , or
is the total el i
E E E E
E E
= Δ + Δ + Δ + ⋅ ⋅ ⋅⋅
=ΣΔ
􀁇 􀁇 􀁇 􀁇
􀁇 􀁇
ectric field. Remember that is a vector that can be resolved
into components, ˆ ˆ ˆ. In the limit where the pieces are small enough,
we can write it as an integral, (or
x y z
E
E Ei E j Ek
E dE
= + +
= ∫
􀁇
􀁇
􀁇 􀁇
, , )
2. Charge Density: when the charges are continuously distributed over a region - a line, the
surface of a material, or inside a sphere - we must specify the
x x y y z z E dE E dE E dE
charge de
=∫ =∫ =∫
. Depending
upon how many dimensions the region has, we define:
(a) For linear charge distribution:
(b) For surface charge distribution:
nsity
dq ds
dq dA
λ
σ
=
=
(c) For volume charge distribution:
The dimensions of , , are determined from the above definitions.
3. As an example of how we work out the electric field coming from a continuous
dq ρ dV
λ σ ρ
=
charge
distribution, let us work out the electric field from a uniform ring of charge at the point P.
2 ( 2 2)
0 0
The small amount of charge gives rise to an electric field whose magnitude is
1
4 4
The component in the z direction is cos z
ds
dE ds ds
r z R
dE dE
λ
λ λ
πε πε
θ
= =
+
= ( )2 21/ 2
with cos z z .
r z R
θ = =
+
( )
( )
( )
( ) ( )
2 23 / 2
0
2 23 / 2 2 23/ 2 2 23 / 2
0 0 0
So . Since , which is the arc length, does not depend upon or ,
4
2
. Answer!!
4 4 4
Note that if you are very far away, t
z
z
dE z ds s z R
z R
E z ds z R qz
z R z R z R
λ
πε
λ λ π
πε πε πε
=
+
= = =
+ + + ∫
( ) 2
0
he ring looks like a point: 1 , .
4
4. As another example, consider a continuous distribution of charges along a wire that lies
along the -axis, as shown below. We want to know the ele
z
E q z R
z
z
πε
= >>
ctric field at a distance from
the wire. By symmetry, the only non-cancelling component lies along the -axis.
x
y
2 2 2
0 0
Applying Coulomb's law to the small amount of charge along the axis gives,
1 1
4 4
the component along the direction is cos . y
dz z
dE dq dz
r y z
y dE dE
λ
λ
πε πε
θ
= =
+
=
2 2
0 0 0
2
Integrating this gives,
cos 2 cos cos .
2
The rest is just technical: to solve the integral, put tan sec . And so,
z z z
y
z z z
E dE dE dE dz
y z
z y dz y d
E
λ
θ θ θ
πε
θ θ θ
λ
=∞ =∞ =∞
=−∞ = =
= = = =
+
= ⇒ =
=
∫ ∫ ∫ ∫
/ 2
0 0 0
cos . Now, we could have equally well taken the x axis. The
2 2
only thing that matters is the distance from the wire, and so the answer is better written as:
d
y y
θ π
θ
λ
θ θ
πε πε
=
=
∫ =
0
.
2
5. The of any vector field is a particularly important concept. It is the measure of the
"flow" or penetration of the field vectors thro
E
r
λ
πε
=
flux
ugh an imaginary fixed surface. So, if there
is a uniform electric field that is normal to a surface of area , the flux is . More
generally, for any surface, we divide the surface up into
A Φ =EA
little pieces and take the
z dE
􀁇
y dE
􀁇
dE
􀁇
y y
r
θ
θ
P
dz dq
z
z
x
component of the electric field normal to each little piece, . If the pieces
are made small enough, then in this limit, .
5. Let us apply the above concept of flux to calc
E i i E A
E dA
Φ = ⋅Δ
Φ = ⋅
Σ
∫
􀁇 􀁇
􀁇 􀁇
2
0
ulate the flux leaving a sphere which has
a charge at its centre. The electric field at any point on the sphere has magnitude equal
to 1 and it is directed radially outwards. Let us now d
4
q
πε r
( 2)
2
0
0
ivide up the surface of the
sphere into small areas. Then 1 4 . So we end up with
4
the important result that the flux leaving this closed surface is .
6. : the
E A E A q r
r
q
π
πε
ε
Φ = Δ = Δ =
Φ =
Σ Σ
Gauss's Law
0
total electric flux leaving a closed surface is equal to the charge
enclosed by the surface divided by . We can express this directly in terms of the
mathematics we have learned, E dA
ε
Φ ≡ ∫ ⋅ 􀁇 􀁇
0
0
. Actually, we have already seen
why this law is equivalent to Coulomb's Law in point 5 above, but let's see it again. So,
applying Gauss's Law to a sphere containing charge,
enclosed q
E d
ε
ε
=
⋅
􀁇 􀁇
( )
0
0
2
0 2
0
. If
the surface is a sphere, then is constant on the surface and and from this
4 1 . This is Coulomb's Law again, but the power of Gauss's
4
law i
enclosed A EdA q
E E dA q
E r q E q
r
ε
ε
ε π
πε
= =
=
= ⇒ =
∫ ∫
∫
s that it holds for any shape of the (closed) surface and for any distribution of charge.
7. Let us apply Gauss's Law to a hollow sphere that has charges only on the surface. At any
distance r from ( 2)
0 the centre, Gauss's Law is 4 . Now, if we are inside the
sphere then 0 and there is no electric field. But if we are outside, then the total
charge is and
enclosed
enclosed
enclosed
E r q
q
q q E
ε π =
=
= = 2
0
1 , which is as if all the charge was concentrated at the
4
centre.
8. Unfortunately, it will not be possible for me to prove Gauss's Law in the short amount of
time and space available bu
q
πε r
t the general method can be outlined as follows: take any
volume and divide it up into little cubes. Each little cube may contain some small amount
of charge. Then show that for each little cube, Gauss's Law follows from Coulomb's Law.
Finally, add up the results. For details, consult any good book on electromagnetism.
+q q =1
r