36
Chapter II Electrostatics Recommended problems: 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 2.14, 2.15, 2.16, 2.20, 2.21, 2.22, 2.23, 2.26, 2.27, 2.32, 2.35, 2.42, 2.43, 2.44, 2.46
Chapter II Electrostatics
Recommended problems: 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 2.14, 2.15, 2.16,
2.20, 2.21, 2.22, 2.23, 2.26, 2.27, 2.32, 2.35, 2.42, 2.43, 2.44, 2.46
The Electric Field
Coulomb's Law
The force acting on a test charge Q due to the source
charge q is given by Coulomb's law as
q
Q
r
F
)1(ˆ4
12r
r
qQF
o
with is the permittivity of free space. 2212 ./1085.8 mNCo
If there are more than one source charge, the net force acting on Q is the
resultant force due to each source charge individually, that is
22
2
212
1
121 ˆˆ
4r
rr
r
qqQFFF
oqi
Q
ri
Fi
ri
r )2(EQFor
is the electric field due to the source charges at point p.
)3(ˆ4
1where
12
n
ii
i
i
o
qE r
r
If the charges are nor discrete but distributed continuously over a region, then
)4(ˆ4
1)(
2 rr
dqrE
o
where dq is the charge of a small line, surface, or volume element, i.e.,
ddqoraddqorlddq ,
Example Find the e.f. a distance z above the midpoint of a
straight line segment of length 2L which carries a uniform line
charge .
r’
dq
dE
r
2L
Solution
)3(ˆ4
1where
12
n
ii
i
i
o
qE r
r
xddq with ixkzrr ˆˆˆand
r
21
2222
ˆˆ
4
1)(
xz
ixkz
xz
xdrEd
o
L
Loz
xz
xdzE
23
224
224
2
Lzz
L
o
0
4 23
22
L
Lox
xz
xdxE
If the rod is very long, i.e., L and noting that
n
n
mx
mn
nmx
0 !1!
!11
zE
oz
2
If the point is far away from the rod, i.e., z>>L
22 44
2
z
Q
z
LE
oo
z
which is the result of a point charge, as expected.
Field Lines
The e.field in a region is represented by an imaginary lines called the e.field
line with the following properties:
(i) They begin from +ve charges and terminate at –ve charges.
(ii) They can never cross.
(iii) Their No. per unit area is proportional to the intensity of the e.f. at the point.
(iv) The direction of the e.f. at a point is tangent to these lines at that point.
Flux and Gauss's Law
The electric flux, E , through a surface S is a measure of number of field lines
passing through S. In other word we write
SE SdE
.
Gauss’s law states that for any closed surface, the e.flux is proportional to the
net e.charge inside that surface, i.e.,
)5(.o
encQSdE
Now, using the divergence theorem dESdE
.
dQencAnd noting that
ddE0
1
Since it is true for any volume
)6(o
E
Differential form of Gauss' law
The Divergence of E
Let us calculate the divergence of E in another way. It is known that
rr
rr
ˆ)(
4
1ˆ
4
1)(
22
drdqrE
oo
drE
o
)(ˆ
4
12r
r )(4
ˆ2
r
rbut
drrrE
o
)(4
4
o
rE
)(
To recover the integral form we integrate the last eq. to get
d
rdE
V oV
)(
Using the divergence theorem we get
o
encQSdE
.
(i) The charge distribution must have a high degree of symmetry
(spherical, cylindrical with infinite length, plane with infinite
extends).
(ii) The Gaussian surface should have the same symmetry as that
of the charge distribution.
(iii) The point at which E is to be evaluated should lie on the
Gaussian surface.
Although Gauss’s law is always true, it is not always useful. For Gauss’s law to
be useful the following conditions should be satisfied:
If the these conditions are satisfied E will be either constant over
the Gaussian surface such that we can pull E out of the integral or
E is perpendicular to the surface such that the integral is zero.
Example Find the e.f outside and inside a
uniformly charged solid sphere of radius R
and charge q.
For the outside region we select a spherical
Gaussian surface of radius r, concentric with
the sphere. Now we have
o
in
qdAE
o
QEA
24 rAbut
2o4 r
QE
Solution r
Gaussian surface
Q a
For the inside region case we choose a
spherical Gaussian surface of radius ra.
To find the charge qin within the Gaussian
surface of volume Vin, we use the fact that
inin Vq
where is the volume charge density. Knowing that
r
Gaussian surface
Q
a
334 rVin
334
anda
Q
3
3
a
Qrqin
Now, applying Gauss’ law we obtain
o
in
qdAE
3o
324
a
QrrE
3o4 a
QrE
Example 24.7 A long cylinder of radius R
carries a charge density proportional to the
distance from the axis =ks , with k is a constant.
Find the e.f outside and inside the cylinder.
+
+
+
+
+
+
+
+
+
+
+
+
dAc
dAb
dAb
h r Solution As a Gaussian surface we select a circular
cylinder of radius r with height h and
coaxial with the line charge.
Since the cylinder has three surfaces, the integral
in Gauss's law has to be split into three parts:
the curved surface, and the two bases.
o
enc
qddd
cbb
AEAEAE
From the symmetry of the system, E is parallel to both bases.
Furthermore, it has a constant magnitude and directed radially
outward at every point on the curved surface of the cylinder.
EA
3
32
0
22
00
)( khRsdsdzdkdsQNowRh
enc
s
kRE
o3
3
+
+
+
+
+
+
+
+
+
+
+
+
dAc
dAb
dAb
h r
To find the e.f. inside the cylinder we choose
the Gaussian surface shown. Now
o
enc
qddd
cbb
AEAEAE
Again E is parallel to both bases and constant
over curved surface of the cylinder.
o
encQshE
2
3
32
0
22
00
)( khrsdsdzdkdsQNowrh
enc
o
krE
3
2
Example 2.4 Find the e.f. due to an
infinite plane that carries a uniform
charge density .
Gaussian surface
Solution
To solve this problem we select as a
Gaussian surface a small cylinder whose
axis is perpendicular to the plane and
whose ends each has an area A.
o
enc
qddd
cbb
AEAEAE
As we do in the previous example we write Gauss’ law as
Aqin
dAb
dAb
dAc
EA
EA
o
2
AEA
o2
E
Example 2.5 Two infinite parallel
planes carry equal but opposite
uniform charge densities . Find
the e.f. in all the three regions.
E-
E+
d + -
Solution
Each plane produces an e.f equal
to o 2
Applying the super position principle
we note that the net e.f. is zero in
the left and in the right regions while
sum up in the middle region.
The net e.f in this regions is
o
EEE
E
The Curl of E
The e.f. due to a point charge at the origin (r=0 ) is rr
qE
o
ˆ4 2
)7(0 E
Now let us prove the same result using Stoke's theorem.
ˆsinˆˆsince drrdrdrld
)8(11
44 2
bao
b
ao
b
a rr
q
r
drqldE
)9(0 ldE
Recalling Stoke's theorem. ldASdA
S
0 E
E is conservative
ArrAA
r
rrr
rABut
r sin
ˆsinˆˆ
sin
12
The Electric Potential
We can write that 0ESince
)10(VE
where V is a scalar function called the electric potential.
b
a
b
aldVldENow
)11( b
a abb
aVVldVldE
Using the fundamental theorem of gradient we get
Comparing Eqs(8) and (11) we obtain
baoba
rr
qVV
11
4)12(
4 r
qV
o
is the e. potential due to a point charge at the origin. Also we have
)13(0)(with)( VldErVr
Example 2.6 Find the potential inside and
outside a spherical shell of radius R, which
carries a uniform surface charge.
r
Q R
Solution
From Gauss's law we have
0,ˆ
4 2 in
oout E
r
rqE
Now for outside region, we have
rout
rldEldErV
)(r
q
r
drqrV
o
r
o 44)(
2
For the inside region, we have
r
R inR
outr
ldEldEldErV
)(
R
q
r
drqrV
o
R
o 44)(
2
Which is constant and equal to the potential on the surface.
Poisson Equation and Laplace's Equation:
It is known that VE
o
Eand
.)14(2 EqPoissonVo
If there is no charge in a region, then we have
.')15(02 EqsLaplaceV
The Potential of a Localized Charge Distribution:
r
qV
o4
The potential of a point charge q at the origin is
In general, the potential of a point charge q is
ro
qV
4
For a collection of point charges we have
)16(4
1
1
n
i i
i
o
qV
r
For a continuous charge distribution we have
)17(4
1r
dqV
o
rr
ldradrdr
ooo
)(
4
1)(
4
1)(
4
1
Example 2.7 Find the potential inside and outside a spherical
shell of radius R, which carries a uniform surface charge.
Solution
Since the charge is distributed over a surface we have
r
adrV
o
)(
4
1
kzrand ˆ
kRjRiRrRrand ˆcosˆsinsinˆcossinˆ
cos222 RzzRrr
r
ddRadNow sin2
022
2
0
2
cos2
sin
4 RzzR
ddRV
o
0
222
cos21
2
RzzRRz
R
o
22
2zRzR
z
RVor
o
RzzRRzRz
zRNow2
RzR
Rzz
R
zV
o
o
2
)(
24 R
q
Knowing that
RzR
q
Rzz
q
zV
o
o
4
4)(
The Boundary Conditions:
Let us study the B.C. at the surface
of two regions in space.
Region I
Region II
Applying Gauss's law to pillbox
shown
o
encQSdE
.
Assuming that the height of pillbox
is negligible (why?)
o
anEanEa
2211 ˆˆ
nnnBut ˆˆˆ 21
n1
n2
)18(21o
nn EE
En is discontinuous at the boundary by o
Region I
Region II
And select the closed loop shown
with the vertical sides are so small
(why?)
0relationtheUsing ldE
02211 lElE
lllBut
21
021 lEE
)19(021 tt EE
Et is continuous at the boundary
The 2-boundary conditions can be combined in a single eq. as
)20(ˆ21 nEEo
Let us calculate the potential across the boundary. We have
2
121 ldEVV
02
121 ldEVV
But since the vertical sides shrinks to zero
)21(21 VV
V is discontinuous across the boundary
Now since VE
Eq.(20) can be written as
nVVo
ˆ21
n
VnV ˆthatfacttheUsing
)22(21
on
V
n
V
Work and Energy
The work required to bring a test charge Q from points a to b is
)23(ab
b
a
b
a
VVQldEQldFW
The work required to bring a test charge Q from points to r is
)24()(rQVW
Let us calculate the work required to assemble a collection of
point charges by bringing them one by one from ∞.
Energy of a Point Charge Distribution
Now, no work is required to place the
first charge q1 at a given position
because there is no electric potential at that position W1=0 .
q1
q2 r12
q3
r23
r13
Next we place a second charge q2 at a position r12 from q1.
This requires a work q2V1 where V1 is the potential at the location of q2 due to q1, or
Next we place q3 at a position r31 from q1 and r32 from q2.
12
212
r
qqkW
We now must do work given as q3V12, where V12 is the potential at
the location of q3 due both q1 and q2, i.e.,
32
2
31
133
r
q
r
qkqW
The net work required to assemble the first 3-charges is
23
32
13
31
12
21
4
1
r
r
r
qqW
o
The total work required to assemble all the system is
ijr
qqW
n
j ij
jin
io
114
1
ijr
qqWor
n
j ij
jin
io
118
1
ijr
n
j ij
j
o
n
ii
1121
4
1
)25()(1
21
n
iji rVqW
Energy of Continuous Charge Distribution
For a continuous charge density Eq.(25) becomes
)26()()()()()()(21
21
21
21
dlrVrdarVrdrVrddqW
Where the integral is to be performed over the charge distribution
o
ENow
drVEW o )(
2
VEEVEVBut
dVEdEVW oo
22
Using the divergence theorem and noting that EV
dESdEVW oo 2
22
If the integral is to be performed over all the space, the 1st integral
vanish (since E 0 as r ∞)
spacethealloverdEW o )27(2
2
Example 2.8: Find the energy of uniformly charged
spherical shell of charge q and radius R.
SolutionI: Using Eq.(26) we have
Q R
darVrddqW )()(21
21
But V on the surface is constant and equal to R
qV
o4
R
qW
o8
2
SolutionII: Using Eq.(27) we have
dEW o 2
2
Rout
oR
ino dEdE
2
0
2
22
0,ˆ
4 2 ino
out Er
rqEBut
R oo
o
R
q
r
drddrqW
8
sin
42
2
4
2
2
2
Conductors
Conductors in electrostatic equilibrium have the following
properties:
(i) Inside the conductors E=0 .
Eo + - + - + - + - + - + - + - + - + - + - + -
E1
If it is not, the free charges would move and it
wouldn’t be electrostatic any more.
Consider a conductor is put into an external e.f. Eo,
this will drive free +ve charge to the left and –ve
charges to the right.
These induced charges a filed E1 opposite to Eo .
These induced charges will continue to flow until the
two fields cancel out such that the net field inside is
zero.
(ii) Inside the conductors =0 .
(iii) The charge resides on the outer surfaces.
(iv) A conductor is an equipotential surface.
(v) Just outside a conductor E is perpendicular to the surface and
equal to /o .
This follow from Gauss’s law o
rE
)(
Since E=0 then =0
If a and b are 2-points within or at the surface of a conductor, then
b
aab ldEVV
But E=0 inside the conductor Vb=Va
Applying the B.C to the surface of a conductor gives
nEEo
inout ˆ
0inEBut
)28(n̂Eo
out
Surface Charge and the Force on a Conductor:
Applying the B.C to the surface of a conductor again
nEEo
inout ˆ
o
inout
n
V
n
V
constantinVBut )29(
n
Vouto
In the presence of an external e.f. the surface charge will experience a force
according to
EAEQF
The force per unit area is )30(Ef
But since is E discontinuous at the surface
nEEEEo
outinoutavr ˆ21
21
21
)31(ˆ2
2
nfo
which is the pressure on the surface of a conductor.
Capacitors:
The capacitor is composed of two conductors separated by an insulator.
When charged the two conductors have equal but opposite charge, the
capacitance of a capacitor is defined as
)32(V
QC
With Q is the charge on either conductor, and V is the potential difference
between the two conductors.
To charge a capacitor electrons are transformed from one plate to the other. In
doing so work is done by an external agent.
This work is stored as potential energy inside the e.f. of the capacitor. This work
can be calculated easily as
)33(221 CQdq
C
qW
Q
o
If the separation d is small compared to the size
of the plates we can assume that any point
inside the capacitor is very closed to a
conductor (just outside a conductor).
Example 2.10: Find the capacitance of a parallel-plate capacitor.
The value of the electric field inside the capacitor is, therefore,
uniform and equal to o
E
dA
QEdldEV
o
b
a
V
QCNow or
oAQd
QC
d
AC o
+Q
-Q
a
b
To find the capacitance of such a capacitor
we have to calculate the potential difference
between the two shells
b
a
dV rE
The e.field between the two shells can be found using Gauss’ law
rE ˆ4 2
or
Q
e. Field between a and b
Note that the e.field outside the capacitor is zero.
Example 2.11: Find the capacitance of two
concentric metal shells with radii a and b.
b
a r
drQV
2o4
b
ar
Q
1
4 o
ab
QV
11
4or
o
From which we have ab
abC
o4
