Download CHAPTER 22 THE ELECTRIC FIELD II CONTINUOUS CHARGE

CHAPTER 22
THE ELECTRIC FIELD
II CONTINUOUS CHARGE DISTRIBUTIONS
!
• Calculating E from Coulomb’s Law
•
•
•
•
•
•
Finite line of charge
Infinite line of charge
Ring of charge
Uniformly charged disk
Uniformly charged infinite plate
Gauss’s Law
•
•
Definition of Electric Flux
!
Calculating E from Gauss’s Law
•
• Outside/inside a spherical conductor
• Infinite line of charge
• Uniformly charged infinite plate
• Uniformly charged insulating sphere
!
E inside hollow conductors
Electric fields are important because the force on a charge
!
!
q is F = qE and so we can determine how a charge moves
in the field, using Newton’s 2nd Law: viz:
!
!
! F qE
a= = .
m m
!
Basically, there are two approaches to calculate E:
•
using Coulomb’s Law
•
using Gauss’s Law
1: Using Coulomb’s Law:
Consider the force on a test charge q " due to an element
of charge dQ within some charge distribution:
dQ
!
r
q"
The force on q " due to dQ is
!
!
dF = q "dE .
!
!
dF 1 q " .dQ
∴dE(at q " ) =
= k 2 ˆr
q" q"
r
i.e.,
!
dQ
E(at q " ) = k ∫ 2 ˆr .
r
Line of charge uniformly distributed
• length 2a, total charge Q, charge per unit length λ
Line of charge uniformly distributed
• length 2a, total charge Q
dy
dy
dQ = λdy
kˆ
r
2a
x
θ
dE x
×
!
E = E x ˆi = k
If x >> a then:
r
ˆj
2a
x
θ
ˆj
dE x
×
ˆi
!
dE
dE y
The y-component of the contribution from the lower
segment cancels the y-component from the upper
x
(x
Q
2
+a
2
)
segment. Thus, the resultant field is confined to the x-
ˆi
direction. So, the magnitude of the electric field along ˆi
due to the segment dy (at y) is
dQ x
xλdy
dE x = dE.cosθ = k 2 . = k
.
2
2 32
r r
x +y
!
Q
E = k 2 ˆi
x
If a ⇒ ∞ then:
(infinitely long wire)
kˆ
ˆi
!
dE
dE y
dQ = λdy
!
E=
(
λ ˆ
i
2πε" x
(
where λ is the charge per unit length =
Q
a
∴E x = kxλ ∫
−a
)
2a .
(x
dy
2
+y
)
,
2 32
)
since k, x and λ are constants. However, the integral is a
standard integral, viz:
⎡
⎤a
y
⎢
⎥
∫
32 = ⎢ 2
2
2⎥
2
2
x
x
+
y
⎣
⎦ −a
−a x + y
a
dy
(
)
⎡
⎤a
y
⎥
∴E x = kxλ ⎢ 2 2
⎢⎣ x x + y 2 ⎥⎦
−a
⎡
⎤
a
(−a)
2aλ
⎥=k
= kxλ ⎢
−
,
⎢⎣ x 2 x 2 + a 2 x 2 x 2 + (−a)2 ⎥⎦
x x2 + a 2
!
E = E x ˆi = k
• Note: when x >> a, then
Q
2
x x +a
2
ˆi .
!
Q
E = k 2 ˆi .
x
• If a → ∞, i.e., an infinite line of charge, then
Q
Q 2kλ
E = Lima →∞ k
=k
=
,
2
2
xa
x
x x +a
i.e.,
! 2kλ
ˆi = λ ˆi .
E=
x
2πε" x
What difference would it make, if any, if the charge was
not distributed uniformly nor symmetrically along the
wire?
On the axis of a line of charge:
• length 2a, total charge Q
On the axis of a line of charge:
• length 2a, total charge Q
y
y
E ˆj
× y
ˆj
y>a
y
d
ˆj
ˆi
y
ˆi
dy
2a
dy
E ˆj
× y
d>a
x
2a
x
The electric field is in the ˆj direction. The magnitude of
E yˆj = k
If y >> a then:
(
Q
ˆ
2
2 j
y −a
Q
E yˆj = k 2 ˆj
y
i.e., wire looks like a point charge.
)
the electric field a distance d from the center of the line
of charge due to the segment dy is
dQ
λdy
dE y = k
.
2 =k
(d − y)
(d − y)2
a
dy
2.
(d
−
y)
−a
∴E y = kλ ∫
Change the variable and put u = (d − y), then dy = −du;
also, when y = ±a, u = d ∓ a.
Question 22.1: If the charge distribution along the wire is
not uniform nor symmetrical, what would be the effect on
the electric field at y?
y
Therefore, we get
!
×E
⎡ 1
du
1 ⎤
=
kλ
=
kλ
−
⎢
⎥
⎢
⎥
2
⎣ u ⎦ d+a
⎣ (d − a) (d + a) ⎦
d+a u
d−a
E = −kλ ∫
⎡ 1 ⎤ d−a
y
⎡ (d + a) − (d − a) ⎤
2kλa
Q
= kλ ⎢
=k 2 2 .
⎥= 2
2
⎣ (d − a)(d + a) ⎦ d − a
d −a
(
!
∴ E = E y ˆj = k
• If d >> a, then
(d
) (
Q
2
−a
2
)
)
x
ˆj.
!
Q
E → k 2 ˆj,
d
i.e., the line of charge looks like a point charge.
A: No affect at all.
B: There would be a component of the field in the xdirection.
C: The field at y is along the y-direction but could be
larger or smaller depending on the exact distribution.
D: Any of the above ... impossible to say without
knowing the precise charge distribution.
!
× E2
!
× E1
y
y
a
2a
x
x
2a
a
Imagine the most extreme cases, i.e., when all the charge
is either at one end or the other (above); the electric
fields are both in the y-direction. However, the strengths
of the fields are different:
!
Q ˆ
E1 = k
j and
(y − a)2
!
E2 = k
Q ˆ
j
(y + a)2
i.e., E1 > E2 . So, the magnitude of the electric field does
depend on how the charge is distributed.
Therefore, the answer is C.
Question 22.2: An infinitely long line of charge that has a
uniform linear charge density equal to −1.50 µC/m lies
parallel to the y-axis at x = −2.00 m. A positive point
charge of magnitude 1.30 µC is located at
( x, y ) = (1.00 m,2.00 m) .
Find the total electric field at ( x, y ) = (2.00 m,1.50 m) .
On the axis of a ring of uniformly distributed charge (Q):
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
y
+
(1,2)
+
P (2,1.5)
!
!
E1
E2
2m
0.5 m
r
26.6"
P
+ +
x
(ˆi )
λ ˆ −1.5 × 10 −6 ˆ
i=
i = −6.75 × 103 ˆi N/C.
2πε" x
2πε" × 4.00
−6
!
q2
9 1.3 × 10
3
E2 = k 2 = 9 × 10 × 2
2 = 9.36 × 10 N/C.
r
0.5 + 1.0
!
!
∴ E2x = E2 cos(26.6" )ˆi = 8.37 × 103 ˆi N/C
!
!
and
E2y = − E2 sin(26.6" ) ˆj = −4.19 × 103 ˆj N/C.
!
∴ E R = (−6.75 ×103 ˆi + 8.37 × 103 ˆi − 4.19 × 103 ˆj) N/C
= (1.62 × 103 ˆi − 4.19 × 103 ˆj) N/C.
+
+
+
kˆ
r = x2 + a 2
a
1.0 m
The total electric field at P is the vector sum
!
!
!
of two fields, i.e., E R = E1 + E2 .
!
E1 =
dQ
P
θ
x
dE x
!
dE
×
dE y
+
ˆj
ˆi
By symmetry, the components parallel to ˆj and kˆ cancel
so the electric field is confined to the x-direction (ˆi ).
The magnitude of the electric field at P due to the
element dQ is
⎛ x ⎞ dQ
xdQ
dE x = cosθdE = ⎜ ⎟ k 2 = k 3 .
⎝ r⎠ r
r
∴E x = k ∫
(x
Q
xdQ
2
+a
=k
(x
)
2 32
Qx
2
since x and a are constant.
+a
=
(x
kx
2
,
2 32
)
+a
∫ dQ
2 32Q
)
On the axis of a ring of uniformly distributed charge (Q):
Q
!
∴ E = E x ˆi = k
(x
Qx
2
+a
• When x = 0:
!
E = 0.
• When x >> a:
!
Q
E = k 2 ˆi
x
+ +
a
+
+
)
2 32
ˆi .
P
×
x
Ex
ˆi
+
+
max at
x= a 2
Ex
Q
E xˆi = k 2 ˆi
x
(x >> a)
+
Ex < 0
(x < 0)
i.e., the ring looks like a point charge.
+
Ex > 0
(x > 0)
E xˆi = k
(x
Qx
2
+a
)
2 3/ 2
ˆi
x
Q
+
+
y
+
z
+
a
+
+
a
×
ˆi
x
!
Ey
y
!
E
×
Q
!
Ex
z
x
x
+
+
+
+
Look at an extreme case, when the charge is concentrated
Question 22.3: If the charge Q is not uniform nor
in one region, at the bottom, say. Clearly, there is a y-
symmetrical around the wire, what difference, if any,
component.
would it make to the electric field at ×?
In general, with a non-uniform charge distribution, the yA: No effect at all.
and z-components of the electric fields from diametrically
B: There would be a components of the field in the y-
opposite sides will not cancel. Therefore, the resultant
and/or z-directions
electric field will not be confined to the x-axis; there will
C: The field at x is along the x-direction but could be
be components in the y- and z-directions, depending on
larger or smaller depending on the distribution
exactly how the charge is distributed. (Non-trivial
D: Any of the above ... not possible to say without
calculation!)
knowing the precise charge distribution
Therefore, the answer is B.
+
Q1
+
Q1
+
+
+
+
+
+
+
+
Q2
+
Q2
+
+
+
1.00 m
+
The magnitude of the electric field at a distance x along
1.00 m
+
Question 22.4: Two uniformly charged rings, one with
charge Q1 and the other with charge Q2, are placed
1.00 m apart with their central axes aligned, as shown
above. The smaller ring, with charge Q1, has radius
a 1 = 0.20 m; the larger ring has radius a 2 = 0.30 m. If
Q1 = 1.50 × 10−6 C, what is Q2 if the electric field
midway between the rings is zero?
the axis of a uniformly charged ring of radius a is
Qx
E=k
32,
x2 + a2
(
)
where Q is the charge on the ring. The electric fields at
x = 0.50 m are directed in opposite directions and so for
E net = 0, we require
Q1x
k
x 2 + a 12
(
i.e.,
)
32
=k
(x
Q2 x
2
+ a2
⎛ x 2 + a 22 ⎞
⎟
Q2 = ⎜ 2
⎝ x + a 12 ⎠
,
2 32
)
32
Q1.
Uniformly charged disk of radius R
(Charge per unit area ⇒ σ = Q 2 )
πR
Charge on ring
+++
kˆ
dQ = 2πrσ.dr
+
+
++
+
+
+ +r ++
+
+
+
+
+
R
+
+
+ + + ++
+ ++
Substituting the given values we obtain
⎛ 0.50m2
32
+ 0.30m2 ⎞
Q2 = ⎜
2
2⎟
⎝ 0.50m + 0.20m ⎠
(1.50 ×10−6 C)
0.34⎞ 3 2
=⎛
1.50 × 10 −6 C = (1.269) 1.50 × 10−6 C
⎝ 0.29⎠
(
)
= 1.90 × 10−6 C.
(
)
x
×
ˆj
dEx
ˆi
⎡
⎤
⎢
⎥
!
σ
1
⎢1 −
⎥ ˆi
E = E x ˆi =
2
⎛
⎞
2ε" ⎢
⎜ R 2 ⎟ + 1 ⎥⎥
⎢⎣
⎝ x ⎠ ⎦
If R >> x, i.e., a very large plate then:
E xˆi
!
σ ˆ
E=
i
2ε"
σ ˆ
i
2ε"
x
−
σ ˆ
i
2ε "
! σ
ΔE = ˆi
ε"
NOTE: when R >> x the field is perpendicular to the plate
and independent of x.
Uniformly charged disk of radius R
• Charge per unit area ⇒ σ = Q
+++
+
+
++
+
+
+ + r+ + +
+
R + + ++
+
+ + + ++
+ ++
πR
Then
2
kˆ
dQ
×
x
ˆj
dEx
ˆi
i.e.,
(
dQ.x
(2πrσ.dr)x
=
k
.
2
2 32
2
2 32
x +r
x +r
)
(
R
∴E x = 2πxkσ ∫
0
(x
)
r.dr
2
+r
x2
2
⎡
⎤
⎢
⎥
σ
1
⎢1 −
⎥,
=
⎛ R2 ⎞ ⎥
2ε! ⎢
⎟
1+ ⎜
⎢⎣
⎝ x 2 ⎠ ⎥⎦
the magnitude of the electric field due to the ring element
dE x = k
∫
2
⎡ −2 ⎤ x +R
du
= πxkσ ⎢ 1 2 ⎥
⎣ u ⎦ x2
u3 2
⎡1
⎤
⎡
⎤
1
x
= 2πxkσ ⎢ − 2
=
2πkσ
1
−
⎥
⎢
⎥
⎣x
⎣
x + R2 ⎦
x2 + R2 ⎦
The ring element has charge dQ = 2πrσ.dr. From before,
is
E x = πxkσ
x2 + R 2
⎡
⎤
⎢
⎥
"
σ
1
ˆ
⎢
⎥ ˆi .
E = Ex i =
1−
⎛ R2 ⎞ ⎥
2ε! ⎢
⎟
1+ ⎜
⎢⎣
⎝ x 2 ⎠ ⎥⎦
• When R >> x, i.e., the plate is very large
.
2 32
)
"
σ ˆ
E→
i
2ε!
We change the variable and put u = x 2 + r 2 . Then
du = 2r.dr, i.e., r.dr = du 2 . Also, when r = 0, u = x 2 and
i.e., the electric field is perpendicular to the plate and
when r = R, u = x 2 + R 2 .
independent of x.
E xˆi
"
σ ˆ
E=
i
2ε!
(R >> x)
−
σ ˆ
i
2ε!
x
" σ
ΔE = ˆi
ε!
σ ˆ
i
2ε !
• When x >> R, i.e., R
2
x2
Electric field due to two large uniformly charged parallel
plates
(Charge per unit area ⇒ σ)
1
+σ
<< 1, we expect the disc
to look like a point charge. When R
2
x2
⎛ ⎛ 2 ⎞ ⎞ −1 2
1−
= 1− ⎜ 1+ ⎜ R 2 ⎟ ⎟
⎛ 2 ⎞
⎝ ⎝ x ⎠⎠
1 + ⎜ R 2⎟
⎝ x ⎠
2
⎛
⎛ 2 ⎞⎞
≈ 1 − ⎜ 1 + −1 2 ⎜ R 2 ⎟ ⎟ = R
.
⎝ x ⎠⎠
⎝
2x 2
( )
But σ =
Q
!
σ ˆ
E1 = −
i
2ε"
2
and ε! = 1 4πk , so
4πR
⎛
⎞
E x = k⎜ Q 2 ⎟ ,
⎝ x ⎠
i.e., the disc looks like a point charge ... good!
!
σ ˆ
E1 =
i
2ε"
1
2
2
ˆi
Consider each plate separately ...
( )⎛⎜⎝ R 2x ⎞⎟⎠ .
∴E x = σ 2ε
!
−σ
d
<< 1, then
1
2
!
σ ˆ
E2 = −
i
2ε"
!
σ ˆ
E2 =
i
2ε"
2
Two large uniformly charged parallel plates ...
1
2
+σ
−σ
d
!
E2
1
2
!
E1
!
E2
A
B
!
!
!
σ ˆ σ ˆ σˆ
In region B: E B = E1 + E2 =
i+
i= i
2ε"
2ε "
ε"
!
!
!
σ ˆ ⎛ σ ⎞ˆ
In region C: EC = E1 + E2 =
i + ⎜−
⎟i = 0
2ε"
⎝ 2ε" ⎠
1
2
! σ
E = ˆi
ε"
NOTE: the field is parallel and
uniform between the plates and
zero outside ... see photo in
chapter 21.
+
C
⎛ σ ⎞
!
!
!
σ ˆ
In region A: E A = E1 + E2 = ⎜ −
i=0
⎟ ˆi +
⎝ 2ε" ⎠ 2ε"
-
σ
!
E1
+
+
+
+
×
+
+
Q
+
+
+
0.50 m
Question 22.5: A uniformly charged ring, with a charge
of 2.50µC and radius 0.20 m, and a very large, uniformly
charged sheet, with charge density 1.00µC/m2 , are
placed 0.50 m apart such that the axis of the ring is
perpendicular to the sheet, as shown above. What is the
electric field at the point × midway between the ring and
the sheet?
σ
+
+
+
+
+
ˆi
+
×
+
Q
+
+
+
0.50 m
Since the charged sheet is very large and positively
charged, the electric field at × due to the sheet is
!
σ ˆ
E1 =
i.
2ε"
1
Since k =
,
4πε!
⎡
⎤
"
1
Qx
⎢σ −
⎥ ˆi .
E net =
2
2 32
2ε ! ⎢
2π x + a
⎥⎦
⎣
(
Substituting the given values:
"
1
−6
E net =
−12 1.00 × 10
2 × 8.854 × 10
[
⎤
2.50 × 10 −6 × 0.25 ˆ
⎥i
−
2
2 32
2π 0.25 + 0.20
⎥⎦
(
The electric field at × due to the positively charged ring
(
is
!
E2 = k
Qx
)
)
= −1.15 × 105 N/C ˆi .
ˆ
3 2 (− i ),
(x2 + a 2 )
where x is the distance from the center of the ring to ×
and a is the radius of the ring.
⎡
⎤
!
!
!
σ
Qx
⎥ˆ
∴ E net = E1 + E2 = ⎢
−k
3 2 i.
2
2
2ε
⎢⎣ "
x +a
⎥⎦
(
)
)
Hence, a positive test charge placed at × will experience
a force directed in the − ˆi direction.
Show for yourself that if Q = 8.24 × 10−7 C (0.824µC)
then the net electric field at × is zero.
E
Imagine an electric field E
passing perpendicularly through
A
an area A. We define the
electric flux:
Φ = EA
!
E
!
A
θ
!
If the electric field E passes
through at an angle θ to the
surface normal. The electric
flux (or simply the flux) is:
! !
Φ = E • A = EA cosθ.
!
Note that area A is actually a vector (directed along the
suface normal) so the flux can be ± depending on θ, i.e.,
!
!
the relative directions of E and A.
If the electric field varies over the surface then the flux
is:
!
!
Φ = ∫ E • dA .
S
!
Question 22.6: A uniform electric field E = aˆi + bˆj
!
intersects a surface of area A. Find the flux through this
area if the surface lies in the
(a) yz plane,
(b) xz plane,
(c) xy plane.
z
2: Using Gauss’s Law:
y
Gauss’s Law: The net flux through any closed 3-d
!
E = aˆi + bˆj
x
surface is:
kˆ
ˆj
(a) surface in the yz plane:
!
ˆi
A = Aˆi .
! !
Φ = E • A = (aˆi + bˆj) • Aˆi = aA
Q
Φ = en
ε!
"
"
(= ∫ E • dA ),
where Qen is the net charge enclosed by the surface.
Gaussian surface
Qen
kˆ
ˆj
(b) surface in the xz plane:
!
ˆi
A = Aˆj.
! !
Φ = E • A = (aˆi + bˆj) • Aˆj = bA.
It is valid for all closed surfaces and all charge
distributions. Flux can flow outward and inward through
a surface; thus, the net flux is the difference between the
kˆ
ˆj
(c) surface in the xy plane:
!
ˆi
A = Akˆ .
! !
Φ = E • A = (aˆi + bˆj) • Akˆ = 0.
outward flux and the inward flux.
Check out a proof of Gauss’s law for an isolated point
charge on the web-page (under the useful notes link).
Take any Gaussian surface enclosing a charge Q. Since
we can take any closed surface, we’ll take a sphere.
Q
+
The total flux Φ through the surface depends on the
number of electric field lines N leaving (or entering) the
charge,
i.e.,
But
Φ ∝ N.
Q
Φ= .
ε!
∴N ∝ Q
i.e., the number of lines leaving or entering a charge Q
is proportional to Q.
Question 22.7: A spherical Gaussian surface is drawn
around a point charge positioned at its center. If the
radius of the sphere is decreased, does the flux increase,
decrease or stay the same?
q
Assume the initial radius is r1 and the
final radius is r2 ( r2 < r1). The flux is
through an element of the surface is
!
!
defined as dΦ = E • dA . But, at every point on a spherical
! !
Gaussian surface E ℓ A and so
!
!
E • dA = E.dA.
+
Q1
−
Q2
Gaussian surface
Thus, the total flux through the surface is
Φ = ∫ E.dA = E ∫ dA = EA,
where A is the area of the surface and E is the value of the
electric field on the surface. If the radius is r, then
q
q
Φ = k 2 .4πr 2 = 4πkq = ,
ε#
r
1
(since k =
, from the previous chapter). So, Φ is
4πε#
independent of r, the radius, i.e., it remains constant. In
fact, we can generalize this result; the flux is independent
of the size of any Gaussian surface.
Therefore, the answer is C.
Note: if two charges of equal magnitude but of opposite
sign, i.e., Q2 = −Q1, are enclosed within a Gaussian
surface, then the net charge enclosed by the Gaussian
surface is zero. Therefore, the net flux through the
surface is zero. Although lines of electric field are
passing through the surface, there are equal numbers
flowing INWARDS and OUTWARDS so the net flux is
zero!
Gauss’s Law can be used to obtain the electric field from
a given charge distribution. It is normally useful only
when there is a good deal of symmetry.
Two charges and a variety of closed surfaces:
Three helpful pieces of additional information are:
Q1
D
Q2
[1] Any excess charge on a conductor resides entirely on
the surface. (Why ?)
By EXCESS charge we mean the net charge!
A
B
C
[2] The electric field at the surface of a conductor is
always perpendicular to the surface. (Next slide.)
What is the net flux through A ?
... through B ?
[3] Choose a Gaussian surface that match the symmetry
... through C ?
of the electric field, so that
... through D ?
(a) the surface is either perpendicular or parallel to
the electric field lines,
(b) the electric field has the same magnitude at each
point on the surface. (See the following examples.)
[2] The electric field is always perpendicular to the
surface of a conductor ...
Electric field due to a charged conductor:
Total
charge
Q +
R
r
+ +
+
+
+ ++
Gaussian
surface
"
E
σ⇒
+ +R+
+
+
+ ++
Q
4πR 2
"
dA E
R
"
A
+Q
Since the excess charge resides on the surface, it doesn’t
If the electric field was at some angle to the surface, there
!
!
would be parallel, E // , and perpendicular, E , components
!
!
of the electric field. The parallel
E
E
!
component would produce a
E //
force on the surface charges and
so they would move! Static
matter whether the sphere is solid or hollow!
OUTSIDE THE SPHERE ( r > R) ...
Use a spherical Gaussian surface centered on the charged
sphere, with surface area 4πr 2 . Gauss’s Law tells us:
"
"
Q
Φ = en = ∫ E • dA = ∫ EdA = E ∫ dA = EA,
ε!
equilibrium, which is established very rapidly, requires that
!
!
there is no parallel component E // , which means E is
since the electric field is constant over the surface.
Φ
Q
1 Q
Q
∴E(r) =
=
=
=
k
( r > R)
A r 4πr 2 ε! 4πε! r 2
r2
perpendicular to the surface at all points.
i.e., it looks just like a point charge!
INSIDE THE SPHERE ...
Total charge
Q
+ + +
+
+
+ ++
Using Gauss’s Law:
E(r) =
Gaussian surface
r<R
Φ Qen
=
=0
A Aε !
r<R
E
Question 22.8: The electric field at the Earth’s surface is
E=k
Q σ
Es = k 2 =
ε!
R
Q
r2
r >R
about 200 N/C and is directed towards the center of the
Earth. What is the charge on the Earth?
(Radius of the Earth = 6400 km.)
r
0
R
Q
Q
At the surface Es = k 2 , but σ =
.
4πR 2
R
1 4πR 2σ σ
∴Es =
.
= .
4πε! R 2
ε!
For any Gaussian surface around the surface of the Earth
Q
we have: Φ = E • A = en .
ε!
Electric field due to an infinitely long charged wire:
"
E
"
Charge per unit
d
A
length = λ
r
Take a spherical Gaussian surface
centered on the Earth, of radius equal
ℓ
to the radius of the Earth (actually just
slightly larger!). Then we have
" "
Φ = E • A = −E × 4πR 2 ,
"
"
as E (inwards) and A (outwards) are anti-parallel.
Qen
∴Φ = −200 × 4π × (6400 × 10 3 ) 2 =
,
8.85 × 10 −12
i.e., Qen = −200 × 4π × (6400 × 10 3 ) 2 × 8.85 ×10 −12
= −9.11 ×105 C.
Which means that the charge on the Earth is negative.
Also, since Qen = ne, that charge corresponds to
Q
−9.11 ×105
n = en =
= 5.69 × 1024
−19
e
−1.60 × 10
excess electrons!
"
E
"
dA
Take a cylindrical Gaussian surface of length ℓ and radius
"
r, centred on the wire. The electric field is radial and E
"
is the same over the curved surface so E through the
ends is zero. Why??
Using Gauss’s Law:
"
"
Q
Φ = en = ∫ E • dA = ∫ EdA = E ∫ dA = EA
ε#
∴E(r) =
Qen
ℓλ
λ
=
=
.
Aε # 2πrℓε # 2πε# r
This is the same result we got using Coulomb’s Law.
Single, infinite sheet of charge:
Single, infinite sheet of charge:
"
E
"
dA
r
++
+ + +
+ + ++ +
++
Qen +
+
+
++
+ ++
+
ℓ
Charge per unit
area = σ.
"
E
Choose a cylindrical Gaussian surface of radius r and
"
length ℓ. Note that E is perpendicular to the sheet and
"
the ends of cylinder, i.e., there is no E through the curved
surface (and so ℓ is not important). From Gauss’s Law:
"
"
Q
Φ = en = ∫ E • dA = ∫ EdA = E ∫ dA = EA
ε#
∴E =
πr 2 σ
Qen
σ
=
=
Aε # 2(πr 2 )ε# 2ε#
Same as the result using Coulomb’s Law.
"
E
"
A
+
+ + +
+ + + +
+
+
+
Qen +
+
+ +
Charge per unit
area = σ
"
A
"
E
+ + + +
+ +
Choose a rectangular box as a Gaussian surface. From
Gauss’s Law the flux through each surface is:
" "
Q
Φ = en = E • A = EA ,
ε!
"
since E is perpendicular to the ends of box and there is
"
no E through the sides.
Total flux: Φ total ⇒ E(2A).
Q
σA
σ
∴E = en =
=
.
2Aε! 2Aε ! 2ε!
Same result as before!
!
dA E
Electric field due to a uniformly charged insulating sphere:
• radius R, Total charge Q.
Uniform volume charge density:
Q
Q
ρ=
=4
+ + +
Vtotal
+ ++ +
πR 3 " +
+ + r
"
+
3
E dA
+ + + + +
+ +++ +
+
r
+ + +
dA
R
Gaussian
surface
(r > R)
+Q
+
R
INSIDE THE SPHERE ( r < R) ...
At radius r, the enclosed charge = ρV, where V is the
It can be shown very easily (prove it for yourselves) that
for r > R
Q
E(r) = k 2
r
(r > R),
i.e., outside the sphere it looks like a point charge.
enclosed volume.
E
4
Qr 3
∴ Enclosed charge ⇒ Qen = ρ πr 3 = 3 .
3
R
Q
Es = k 2
R
Using Gauss’s Law:
"
"
Q
Φ = en = ∫ E • dA = ∫ EdA = E ∫ dA = EA.
ε!
Qr
E(r) = k 3
R
∴E(r) =
i.e.,
!
A
E(r) =
Qr 3
2
4πr ε !R
r
0
3,
1 Qr
Qr
=
k
.
4πε! R 3
R3
Q
E(r) = k 2
r
(r < R)
R
NOTE: The electric field at the surface of an insulating
Q
sphere is the same as a conducting sphere, Es = k 2 .
R
Question 22.9: A nonconducting sphere of radius
6.00 cm has a uniform volume charge density of
450 nC/m 3.
(a) What is the total charge on the sphere?
Find the electric field at the following distances from the
center of the sphere:
4
(a) Total charge ⇒ Vtot ρ = πR 3ρ
3
4
= π(0.06)3 × 450 ×10 −9 = 4.07 × 10 −10 C.
3
Q
6.0cm
(b) Φ(r) = E(r).A r = en .
ε!
r
ρ = 450 × 10 −9 C m3
Qen (4 3)πr 3ρ rρ
∴E(r) =
=
=
.
A r ε!
3ε!
4πr 2ε!
With r = 2.00 cm ⇒ E(r) = 339 N/C.
(c) With r = 5.90 cm ⇒ E(r) = 999 N/C
(b) 2.00 cm,
(d) When r > R, Qen = 4.07 ×10 −10 C.
(c) 5.90 cm,
Therefore, with r = 6.10 cm,
(d) 6.10 cm, and
(e) 10.0 cm.
E(r) =
Qen 4.07 × 10 −10
=
= 983 N/C.
A r ε!
4πr 2 ε!
(e) With r = 10.0 cm ⇒ E(r) = 366 N/C.
Note: (d) and (e) are the same for a conducting sphere
(solid or hollow) with a total charge of 4.07 × 10 −10 C.
If there is a cavity within the conductor that contains no
The charge on a conductor resides entirely on the surface
but what if the conductor has a hollow cavity?
+
+
+
+
+
+
+
+
+
+ ++
+
+
+
+
Let’s start with a solid conductor with positive charges
distributed over its surface.
outside the cavity.
+
+
+ +
charge, draw a Gaussian surface within the conductor but
+
+
+
+
+
+
+
+
+
+ +
+
+ +
+
+
+
+
From earlier we found that the electric field strength
E = 0 everywhere within a conductor, i.e., there is no net
flux within a conductor. Therefore, Gauss’s Law tells us
that there is no charge within the Gaussian surface, i.e.,
there is no charge on the surface of the cavity.
But, what if the cavity contains a charge +q?
+
+q
+
+
- +++q++ - + +-+
+
E=0
+
−q
+
Question 22.10:
+
R 2 = 2.00 m
R 1 = 1.00 m
Everywhere within a conductor E = 0, so Gauss’s Law
tells us that there must be no net charge inside the
Gaussian surface. So, there must be a charge of −q over
the inner surface of the cavity. If the conductor were
uncharged initially, then the −q charge had to come from
somewhere ... by redistributing charge in the conductor so
the outer surface has a charge +q.
If the conductor had an existing charge +Q initially, then
inserting a charge of +q in the cavity will produce a
charge of −q over the inside of the cavity and (q + Q) on
the outside of the conductor.
The inner conducting sphere, of radius 1.00 m, has a
surface charge density of 1.00 µC/m 2 . What is the
charge density on the inner surface of the outer sphere
(inner radius = 2.00 m)?
R 2 = 2.00 m
R 1 = 1.00 m
How come the electric field is zero inside a conductor
even if there’s an external field? ... because the charges
inside are re-distributed to produce a “compensating”
field, Ecomp, ...
If the total charge on the inner sphere is +Q, then from our
σL
notes, the total charge on the inner surface of the outer
sphere is −Q. So, if A1 is the surface area of the inner
sphere and A2 is the inner surface area of the outer
sphere:
A2σ2 = −A1σ1.
A
4πR12
∴σ2 = − 1 σ1 = −
σ1
A2
4πR 22
=−
2
(1.00 m )
σ = −0.25 µC/m 2 .
( 2.00 m )2 1
Since the charge on the outer surface is +Q, the charge
density on the outer surface is:
A
A
R2
σ3 = − 2 σ2 = 1 σ1 = 12 σ1,
A3
A3
R3
where R 3 is the outer radius of the outer sphere.
-
+
+
-
+
-
+
σR
"
E
Einside = Ecomp − E = 0.
σ
If Einside = 0 then Ecomp = .
ε!
∴E =
σ
, so the charge on
ε!
Initially uncharged
∴σ R ⇒ +σ
and
σ L ⇒ −σ
"
E
−σ
-
+ +σ
-
+
-
"
Ec
+
the surfaces of the conductor is ±σ = ±ε!E, where E
is the external field.
What about a hollow conducting box in an electric
field? ... why is the field inside zero?
Examples of a Faraday cage ...
... coaxial (antenna) wire.
σL + σR = 0
Ecomp =
- Ein = 0 +
- σL σR +
+
-
σ
= E.
−σ ε"
-
∴σ = ε"E
-
!
Ecomp
copper
wire
!
E
+ +σ
+
+
If an uncharged conducting hollow box is placed in an
!
electric field ( E) the charges within the conductor are
redistributed. But, the electric field created by the
!
redistributed charge ( Ecomp) is such that the net field
inside ( Ein) is zero. This configuration is known as a
FARADAY CAGE and is a way of “protecting” sensitive
equipment.
insulation
copper
braid
!
E
Ein = 0
The copper braid acts as a Faraday cage. Since the
electric field inside the braid is zero the center conducting
wire is “protected” from extraneous, external signals.
More examples of Faraday cages!
A large spark jumps to the car’s body and then jumps
across the car’s front tire. The person inside is
unharmed because there is no electric field inside the
vehicle.
Would you volunteer??
!
Earlier we saw that E is always perpendicular to the
surface of a charged conductor. What happens if a
neutral conductor is placed in an electric field?
Since we are conductors, we “distort” the electric field
of the Earth ...
Conductor
E ~ 200N/C
The charges within the conductor are redistributed by the
electric field. However, the resultant electric field - due
to the field created by the conductor and the original field
- is still perpendicular to the surface of the conductor.
Note that the electric field lines are perpendicular to
So, no matter whether it is charged or not, the electric
the body at all points!
field at the surface of a conductor is always
perpendicular to the surface. Of course, the electric field
inside the conductor is zero.
ℓ
σL
E = σ L 2ε "
σR
E = σ L 2ε "
ℓ
ER
EL
ΦL
ΦR
ℓ
2.50 nC/m 2
ℓ
7.50 nC/m 2
ℓ
σL
ℓ
σR
E = σ R 2ε "
E = σ R 2ε "
The electric field due to the σ L plate is E = σ L 2ε " and
the electric field due to the σ R plate is E = σ R 2ε " .
Thus, the total electric field towards the left is:
Question 22.11: A cylindrical Gaussian surface encloses
a region perpendicular to two, very large parallel plates
with different charge densities, σ L = 2.50 nC/m 2 and
σ R = 7.50 nC/m 2 . If Φ L and Φ R are the fluxes through
the left and right ends of the cylinder, what are Φ L and
ΦR?
(
)
E L = σ L 2ε " + σ R 2ε" = 10.0 nC/m 2 ε " ,
and the total electric field towards the right is:
(
)
E R = σ R 2ε " + σ L 2ε" = 10.0 nC/m 2 ε " .
The areas (A) of the ends of the cyclinder are equal, and
the electric fields are constant over the ends, so
Φ L = E L A and Φ R = E R A.
But E L = E R ,
∴Φ L = ΦR .