coulomb’s law point charge :. line charge : surface charge :
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Coulomb’s Law
Point Charge :
30
)(
4)(
rr
rrqrE
r
r
)( rr
Line Charge :
rr
dl
ldrrr
rrrE
)(
)(
4
1)( 3
0
Surface Charge :
rr
)( rr
da
adrrr
rrrE
)(
)(
4
1)( 3
0
dr
rr
rrrE )(
)(
4
1)( 3
0
rr
)( rr
d
Volume Charge
Prob. 2.6: Electric Field at a height z above the centre of a circular plate of uniform charge density.
),0,0( z
0R
kzr ˆ
jrirr ˆsinˆcos
drdr
rz
jrirkzrE
R2
0 02/322
0
0
)(
ˆsinˆcosˆ
4)(
r
r
kRzz
zzE ˆ
)(
11
2),0,0(
2/1220
0
Limiting Case : R (Infinite Sheet)
0ˆ2
),,(0
0 zkzyxE
)0(ˆ2 0
0 zk
Prob. 2.41. Electric Field at a height z above the centre of a square plate of uniform charge density.
dxdydq 0
r
r
dxdydq 0
r
r
jyixrkzr ˆˆ;ˆ
2/
2/
2/
2/2/3222
0
0
)(
)ˆˆˆ(
4)(
a
a
a
a
ydxdyxz
jyixkzrE
Any kind of charge distribution can be treated as a volume distribution
1. Point charges :
....)()()( 23
213
1 rrqrrqr
1r
2r
3r
1q
2q
3q
2. Line Charge
)(z
),()(),,( 2 yxzzyx
3. Surface Charge
)(),(),,( zyxzyx
),(: yxdensitySurface
inityinfinityinf
inityinf
inityinf
Divergence of the Electric Field
dr
rr
rrrE )(
)(
4
1)( 3
0
dr
rr
rrrE )(
)(
4
1)( 3
0
)(4)(
, 33 rr
rr
rrHowever
)(1
)()(1
)(0
3
0
rdrrrrE
Examples :
1. Point Charge
30
)(
4)(
rr
rrqrE
0
33
0
)()(4
4
rrq
rrq
E
00
0 )(
z
z
EE z
1)(22
,0,00
0 zEEE zyx
Infinite Charge Sheet
Gauss’ Law
S V
en
V
qddEadE
00
1)(
1q
2q3q 4q
Ead
SV
43 qqqen
Applications of Gauss’ Law
Although Gauss’ Law is valid for any kind of charge distribution and any Gaussian Surface, its applicability to determine the electric field is restricted only to symmetrical charge distribution
1. Electric field of a point charge
?E
?E
Gaussian Surface
Gaussian Surface
0
24
qErdaEadE
S S
20
ˆ
4 r
rqE
Prob. 2.16 Long co-axial cable : i) Inner solid cylinder of radius a carrying uniform volume charge density ρ ii) outer cylindrical surface of radius b carrying equal and opposite charge of uniform density σ.
Find field in regions i) s<a ii) a<s<b iii) s>b
E
s < a
ab
s
h
E
a < s < bsh
E
s > as
h
ass
sE 02
)(
bsas
sa
ˆ
2 0
2
bs 0
s
)(sE
a b
Prob. 2.17 Infinite plane slab of thickness 2d (-d<y<d) of uniform volume charge density ρ. Find E in regions i) y<-d ii) –d<y<d iii) y>d
d2
E
y2a
-d < y < d
E
y2
dy
Curl of the Electric Field
dr
rr
rrrE )(
)(
4
1)( 3
0
dr
rr
rrrE )(
)(
4
1)(
30
0)(
,3
rr
rrHowever
0
E
pathoftindependenisldEP
P 2
1
VEtsVfieldscalara
..
Where, V can be constructed as :
r
r
ldErV
0
)(
The scalar field is called the electric potential and the point is the zero of the potential
)(rV
0r
Changing the zero of the potential
)(&)( rVrV Let be the potentials with the
zero (reference points) at respectively
00 & rr
0
0
)()(r
r
rVldErV
)()( 0 rVrV
VVE
)()()()( 2121 rVrVrVrV
Prob. 2.20
One of the following is an impossible electrostatic field. Which one?
]ˆ3ˆ2ˆ[) kxzjyzixykEa
]ˆ2ˆ)2(ˆ[) 22 kyzjzxyiykEb
For the possible one, find the potential and show that it gives the correct field
Conventionally, the zero of the potential is taken at infinity :
r
ldErV
)(
Potential of a point charge (zero at infinity) :
r
ldr
rqrV
3
04)(
r
from
rrdld ˆ
r
r
q
r
rdqrV
1
44)(
02
0
Charge located at :r
rr
qrV
1
4)(
0
Potentials of Extended Charge Distributions :
Line Charge :
rr
ldrrV
)(
4
1)(
0
Surface Charge :
rr
adrrV
)(
4
1)(
0
Volume Charge :
rr
drrV
)(
4
1)(
0
Prob. 2.26
The conical surface has uniform charge density σ. Find p.d between points a & b
h
h
a
b
2
drdrad
r
r
2
dr
Prob. 2.9
Suppose the electric field in some region is found to be : rkrE ˆ3
a) Find the charge distribution that could produce this field
b) Find the total charge contained in a sphere of radius R centered on the origin. Do it in two different ways.
)(1 22 rErrr
E
Work done in moving a charge in an electric field
a
b
Eld
EqF
b
a
b
a
ab aVbVqldEqldFW )()(
If the charge is brought from infinity to the point :
)()]()([ rVqVrVqW
r
If V is the potential of a point charge Q located at then,r
rr
QqW
04
1
Q qrr
Electrostatic Energy of a Charge Distribution
It is the work done to assemble the charge configuration, starting from some initial configuration
Initial Config.Given Config.
The standard initial configuration is taken to be one in which all small (infinitesimal) pieces of charge are infinitely separated from one another.
1. Point Charges
1q
2q
Nq1r
2r
32
32
31
31
03
21
21
021 4
1;
4
1;0
rr
rr
qqW
rr
qqWW
1
104
1 i
j ji
jii
rr
qqW
N
i
N
i
i
j ji
jii
rr
qqWW
1 1
1
104
1
N
i
N
j ji
ji
rr
1 104
1
ij
N
i
N
j ji
ji
rr
1 104
1
2
1
ij
N
iii rVq
1
)(2
1
Electrostatic Energy of Continuous Charge Distribution :
V
drVrW )()(2
1
r
d
The electrostatic energy of a charge distribution can be expressed as an integral over the electric field of the distribution :
spaceall
dEW 20
2
Proof :
V
drVrW )()(2
1
V
dEV )(
20
)()()( VEEVEV
2)( EEV
S V
dEadEVW 20
2)(
SV
Including more and more volume in the integral,
spaceall
dEW 20
2
Prob. 2.45
A sphere of radius R carries a charge density . Find the energy of the configuration in two different ways.
krr )(
a) Find the energy by integrating over the field
b) Find the potential everywhere and do the integral : drVrW )()(
2
1
Prob. 2.33
Find the electrostatic energy of a uniformly charged solid sphere of total charge Q by the following method : Calculate work done in adding charge layer by layer
R
rdr
Electrostatic energy of a point charge Q
Energy of a uniform solid sphere of radius R and total charge Q :
R
QWsphere
1
20
3
0
2
Energy of a point charge Q :
sphere
Rpo WW
0int lim
Self and Interaction Energy
1 2
21 & EE
are the fields produced by 21 &
21
2
2
2
1
2
21 2& EEEEEEEE
,,int21 whereWWWW
spaceall
spaceall
dEWdEW
2
20
2
2
10
1 2,
2
spaceall
dEEW 210int
21 &WW are the energies of the two charge distributions, existing alone. They are called the self energies of the distributions
intW is the energy of interaction between them. It is the work done to bring them, already made, from infinity.
Electrostatic Boundary Conditions
1
2
1E2E
n̂
Applying Gauss’ law to the pillbox :
012 )ˆˆ(
S
SnEnE side
As the two flat faces come infinitesimally close to the charged surface, 0side
)1(0
12
EE
Taking the line integral of around the closed loop :
E
)2(012 IIII EE
2E
1E
Combining (1) & (2) : nEE ˆ0
12
Examples :
kEE ˆ0
12
1
2kE ˆ
2 02
kE ˆ2 0
1
1. Infinite Sheet
ass
sE 02
)(
bsas
sa
ˆ
2 0
2
bs 0
sb
aEE bb ˆ
2 0
2
n̂
0
a
b
2. Co-axial Cable :
Conductors
A perfect conductor is a body possessing unlimited supply of charges of each kind (+ve & -ve), at least one of which kind is completely free to move within the body and on its surface.
Mathematically a conductor is capable of developing any charge density with the only constraint :
V
d 0
(Neutral Cond.)
Or, since charge can reside only on the surface of a conductor, the only restriction on the surface charge density is :
0)( darS
+++
+
++
+++++
++
+
++
+++
++------------
--
---
--
-
-
E
Properties of a perfect conductor
1. The electric field within the body of the conductor is zero
At equilibrium (After charge flow in the conductor has ceased) :
Otherwise, there is no reason why charge should stop flowing
+++
+
++
+++++
++
+
++
+++
++------------
--
---
--
-
-
E
i) Does the conductor have the necessary ammunition to nullify the external field within?
ii) How long does it take the conductor to nullify the external field?
2. There cannot be any charge density within the body of the conductor
Gaussian surface S
000 S
enQadE
3. The surface of a conductor is an equipotential surface
a
b
b
a
ldEaVbV 0)()(
)()( bVaV
4. The electric field just outside the surface of a conductor is everywhere perpendicular to the surface
Reason : The gradient is everywhere perpendicular the level surface.
5. The electric field at any point just outside the surface of a conductor is related to the surface charge density at that point by :
):ˆ(ˆ0
normalunitoutwardnnEout
Reason :
From boundary condition on the field :
0ˆ0
ininout EandnEE
Electrostatic Pressure
outE
da
?)( outEdaFd
?)( inEdaFd
or
Answer :
)()(2
1outin EEdaFd
ndaˆ
2 0
2
0
2
2
da
dFP
Note : The surface of the conductor is everywhere pushed outwards.
outE
da
otherselfout EEE
Correctly stated : otherEdaFd)(
nEandnE outself ˆˆ2 00
2out
selfother
EEE
Reason that nEself ˆ2 0
Electrostatic Pressure :
0
2
2
P
Prob. 2.38
A metal sphere of radius R carries a total charge Q. What is the force of repulsion between the two halves of the sphere?
1 10
2
2S S
adadPF
kR ˆ)(2
2
0
2
k̂
1S
2S
Poisson and Laplace Equation
VEE
&0
Combining,
)'(0
2 EquationsPoissonV
In a charge free region :
)'(02 equationsLaplaceV
Example (Point Charge) :
r
qrV
1
4)(
0
r
qV
1
42
0
2
However,
)(4ˆ11 32
2 rr
r
rr
0
int3
0
2 )(
por
qV
Prob. 2.46 : The electric potential of some charge configuration is given by :
r
eArV
r
)(
Find the charge density and the total charge Q
Ans:
r
eA
r
20
re
ree
rr
e rrrr 1
211 222
rer
r 2
3 )(4
re
rrA
23
0 )(4
0
20 44 drreAQ r
0
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