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ELECTROSTATICS
Gauss’s Law and Applications
Though Coulomb’s law is fundamental, one finds it cumbersome to use it to cal-
culate electric field due to a continuous charge distribution because the integrals
involved can be quite difficult. An alternative but completely equivalent formula-
tion is Gauss’s Law which is very useful in situations which exhibit certain sym-
metry.
Electric Lines of Force :
Electric lines of force (also known as field lines) is a pictorial representation of the
electric field. These consist of directed lines indicating the direction of electric
field at various points in space.
• There is no rule as to how many lines are to be shown. However, it is
customary to draw number of lines proportional to the charge. Thus if N
number of lines are drawn from or into a charge Q, 2N number of lines
would be drawn for charge 2Q.
• The electric field at a point is directed along the tangent to the field lines.
A positive charge at this point will move along the tangent in a direction
indicated by the arrow.
• Lines are dense close to a source of the electric field and become sparse as
one moves away.
• Lines originate from a positive charge and end either on a negative charge
or move to infinity.
1
• Lines of force due to a solitary negative charge is assumed to start at infinity
and end at the negative charge.
• Field lines do not cross each other. ( if they did, the field at the point of
crossing will not be uniquely defined.)
• A neutral point is a point at which field strength is zero. This occurs because
of cancellation of electric field at such a point due to multiple charges.
Exercise : Draw field lines and show the neutral point for a charge +4Q located
at (1, 0) and −Q located at (−1, 0).
2.3 Electric Flux
The concept of flux is borrowed from flow of water through a surface. The amount
of water flowing through a surface depends on the velocity of water, the area of the
surface and the orientation of the surface with respect to the direction of velocity
of water.
Though an area is generally considered as a scalar, an element of area may be
considered to be a vector because :
• It has magnitude (measured in m2
).
• If the area is infinitisimally small, it can be considered to be in a plane. We
can then associate a direction with it. For instance, if the area element lies
in the x-y plane, it can be considered to be directed along the z–direction.
(Conventionally, the direction of the area is taken to be along the outward
normal.)
2
S
θ
In the figure above, the length of the vector S is chosen to represent the area
in some convenient unit and its direction is taken to be along the outward
normal to the area.
We define the flux of the electric field through an area dS to be given by the scalar
product
dφ = E · dS
If θ is the angle between the electric field and the area vector
dφ =| E || dS | cos θ
. For an arbitrary surface S, the flux is obtainted by integrating over all the surface
elements
φ = dφ =
S
E · dS
If the electric field is uniform, the angle θ is constant and we have
φ = ES cos θ = E(S cos θ)
Thus the flux is equal to the product of magnitude of the electric field and the
projection of area perpendicular to the field.
3
S
E
θ
Unit of flux is N-m2
/C. Flux is positive if the field lines come out of the surface
and is negative if they go into it.
Solid Angle :
The concept of solid angle is a natural extension of a plane angle to three dimen-
sions. Consider an area element dS at a distance r from a point P. Let ˆn be the
unit vector along the outward normal to dS.
The element of the solid angle sub-
tended by the area element at P is de-
fined as
dΩ =
dS⊥
r2
where dS⊥ is the projection of dS
along a direction perpendicular to r.
If α is the angle between ˆr and ˆn,
then,
dΩ =
dS cos α
r2
r
n^
^
P d Ω
α
dS
Solid angle is dimensionless. However, for practical reasons it is measured in
terms of a unit called steradian (much like the way a planar angle is measured in
terms of degrees).
The maximum possible value of solid angle is 4π, which is the angle subtended
by an area which encloses the point P completely.
Example 1:
A right circular cone has a semi-vertical angle α. Calculate the solid angle at the
apex P of the cone.
Solution :
The cap on the cone is a part of a sphere of radius R, the slant length of the cone.
Using spherical polar coordinates, an area element on the cap is R2
sin θdθdφ,
where θ is the polar angle and φ is the azimuthal angle. Here, φ goes from 0 to 2π
4
while θ goes from 0 to α.
Thus the area of the cap is
dA = 2πR2
α
0
sin θdθ
= 2πR2
(1 − cos α)
Thus the solid angle at P is
dΩ =
dA
R2
= 2π(1 − cos α)
R
P θ
Exercise :
Calculate the solid angle subtended by an octant of a sphere at the centre of the
sphere. (Ans. π/2)
The flux per unit solid angle is known as the intensity.
Example 2
An wedge in the shape of a rectangular box is kept on a horizontal floor. The
two triangular faces and the rectangular face ABFE are in the vertical plane. The
electric field is horizontal, has a magnitude 8 × 104
N/C and enters the wedge
through the face ABFE, as shown. Calculate the flux through each of the faces
and through the entire surface of the wedge.
000000000000000000000000
000000000000000000000000
000000000000000000000000000000000000
111111111111111111111111
111111111111111111111111
111111111111111111111111111111111111
0.4 m
0.3m
0.2m
E
A
B
C
DE
F
Solution :
The outward normals to the triangular faces AED, BFC, as well as the normal to
the base are perpendicular to E. Hence the flux through each of these faces is zero.
The vertical rectangular face ABFE has an area 0.06 m2
. The outward normal to
this face is perpendicular to the electric field. The flux is entering through this
5
face and is negative. Thus flux through ABFE is
φ1 = −0.06 × 8 × 104
= −.48 × 104
N − m2
/C
To find the flux through the slanted face, we need the angle that the normal to this
face makes with the horizontal electric field. Since the electric field is perpen-
dicular to the side ABFE, this angle is equal to the angle between AE and AD,
which is cos−1
(.3/.5). The area of the slanted face ABCD is 0.1 m2
. Thus the
flux through ABCD is
φ2 = 0.1 × 8 × 104
× (.3/.5) = +0.48 × 104
N − m2
/C
The flux through the entire surface of the wedge is φ1 + φ2 = 0.
Example 3:
Calculate the flux through the base of the cone of radius R.
E
Solution :
The flux entering is perpendicular to the base. Since the outward normal to the cir-
cular base is in the opposite sense, the flux is negative and is equal to the product
of the magnitude of the field and the area of the base, The flux, therefore is, πR2
E.
Example 4 :
Calculate the flux coming out through the curved surface of the cone in the above
example.
Solution :
6
dl
h
H r
θ
R
Consider a circular strip of radius r at a depth h from the apex of the cone. The
angle between the electric field through the strip and the vector dS is π −θ, where
θ is the semi-angle of the cone. If dl is the length element along the slope, the area
of the strip is 2πrdl. Thus,
E · dS = 2πrdl | E | sin θ
We have, l = h/ cos θ, so that dl = dh/ cos θ. Further, r = h tan θ Substituting,
we get
E · dS = 2πh tan2
θ | E | dh
Integrating from h = 0 to h = H, the height of the cone, the outward flux is
| E | H2
tan2
θ = πR2
| E |.
Example 5 :
A charge Q is located at the center of a sphere of radius R. Calculate the flux
going out through the surface of the sphere.
dS
By Coulomb’s law, the field due to the charge Q is radial and is given on the
surface of the sphere by,
E =
1
4πǫ0
Q
R2
ˆr
7
The direction of the area vector dS, is also radial at each point of the surface
dS = dSˆr. The flux
φ = E · dS
=
1
4πǫ0
Q
R2
dS
The integral over dS is equal to the surface area of the sphere, which is, 4πR2
.
Thus the flux out of the surface of the sphere is
φ =
1
4πǫ0
Q
R2
· 4πR2
=
Q
ǫ0
Note that the flux is independent of the radius of the sphere - a cancellation
due to the fact that the surface area of the sphere is proportional to r2
while
the field is proportional to 1/r2
. Curiously, the result is valid for any arbitrar-
ily shaped surface. Consider a cone of solid angle dΩ centered at the charge.
This will intersect the arbitrarily
shaped surface in an ellipse whose
normal ˆn makes an angle θ with the
outgoing radial direction ˆr. If the
area of the ellipse is dS,
dΩ =
dS cos θ
r2
so that the flux through the cone is
dΦ =
1
4πǫ0
q
r2
ˆr · dS
=
1
4πǫ0
q
r2
ˆr ·
r2
dΩ
cos θ
ˆn
=
1
4πǫ0
qdΩ
ˆn · r
cos θ
=
qdΩ
4πǫ0
(1)
q
dS
r
n^
^
r
dΩ
Thus the total flux throgh the surface is
Φ = dΦ =
q
4πǫ0
dΩ =
q
ǫ0
(2)
8
One can generalize this to multiple charges since superposition principle holds for
the electric field. Suppose the total field consists of fields E1 due to charge q1, E2
due to q2 and so on. We have
E · dS =
i
Ei · dS =
i
qi
ǫ0
One should note that the above
derivation is valid only if the charges
are contained within the volume for
only then the total solid angle be-
comes 4π. For charges which are
outside the volume, the flux that en-
ters the volume also leaves it and
though the areas which the cone in-
tersects are different, the solid angles
are the same. This leads to cancella-
tion of the flux and the contribution
to the flux from a charge which re-
sides outside the volume is zero.
r
dΩ
q
r^
n^
n^ dS
dS1
1
2
2
GAUSS’S LAW - Integral form
The flux calculation done in Example 4 above is a general result for flux out of
any closed surface, known as Gauss’s law.
Total outward electric flux φ through a closed surface S is equal to 1/ǫ0 times
the charge enclosed by the volume defined by the surface S
9
01E
01E
01E
00001111
01dS
Mathematicaly, the surface integral of the electric field over any closed surface is
equal to the net charge enclosed divided by ǫ0
E · dS =
Qenclosed
ǫ0
(3)
• The law is valid for arbitry shaped surface, real or imaginary.
• Its physical content is the same as that of Coulomb’s law.
• In practice, it allows evaluation of electric field in many practical situa-
tions by forming imagined surfaces which exploit symmetry of the problem.
Such surfaces are called Gaussian surfaces.
GAUSS’S LAW - Differential form
The integral form of Gauss’s law can be converted to a differential form by
using the divergence theorem. If V is the volume enclosed by the surface S,
S
E · dS =
V
∇ · Edv (A)
If ρ is the volume charge density,
Q =
V
ρdv (B)
10
Thus we have
∇ · E =
ρ
ǫ0
(4)
Direct Calculation of divergence from Coulomb’s Law :
We will use the field expression
E(r) =
1
4πǫ0
(r − r′
)
| r − r′ |3
ρ(r′
)d3
r
to directly evaluate the divergence of the electric field. Since the differentiation
is with respect to r while the integration is with respect to r′
, we can take the
divergence inside the integral,
∇ · E(r) =
1
4πǫ0
∇r ·
(r − r′
)
| r − r′ |3
ρ(r′
)d3
r
A simple minded calculation of divergence is as follows :
∇r ·
(r − r′
)
| r − r′ |3
= (∇r · r)
1
| r − r′ |3
+ (r − r′
) · ∇r
1
| r − r′ |3
Divergence of r is equal to 3
∇r · r =
∂
∂x
x +
∂
∂y
y +
∂
∂z
z = 3
∇r
1
| r − r′ |3
= −3
r − r′
| r − r′ |5
Thus it would seem that
∇r ·
(r − r′
)
| r − r′ |3
= 0
i.e. ∇ · E = 0, which violates Gauss’s Law, that ∇ · E = ρ/ǫ0. The problem
arises because the function 1/ | r − r′
| has a singularity at r = r′
. This point has
to be taken care of with care. Except at this point the divergence of the integral is
indeed zero.
11
Thus we can shrink the range of in-
tegral till it becomes a small sphere
around the point r(x, y, z). Since
ρ(x′
, y′
, z′
) is continuous, we may
replace the density on the surface of
the small sphere by the value of den-
sity at the centre, so that the density
term can come out of the integral,
leaving,
∇·E(r) =
1
4πǫ0
ρ(r) ∇r·
(r − r′
)
| r − r′ |3
d3
r
r − r’
(x,y,z)
(x’,y’,z’)
Since the divergence with respect to r is being taken of a function which only
depends on the difference r − r′
, we may replace ∇r → −∇′
r, which gives
∇ · E(r) = −
1
4πǫ0
ρ(r) ∇r′ ·
(r − r′
)
| r − r′ |3
d3
r
The volume integral on the right may be converted to a surface integral using the
divergence theore,
∇ · E(r) = −
1
4πǫ0
ρ(r)
(r − r′
)
| r − r′ |3
· dS
where the integral is over the surface of the sphere. If the radius of the sphere be
taken as r0, r − r′
= −r0ˆn and dS = dSˆn, Hence
∇ · E = +
1
4πǫ0
ρ(r)
dS
r2
0
=
1
4πǫ0
ρ(r) · 4πr2
0 =
ρ(r)
ǫ0
Example 1: A sphere of radius R contains a continuous charge distribution ρ(r).
The electric field at a distance r from the centre of the sphere is E = kr3
ˆr.
(a) Find the charge density.
Soln. :
∇E =
1
r2
∂
∂r
(r2
Er) =
1
r2
∂
∂r
(kr5
) = 5kr2
= ρ/ǫo
Thus ρ = 5kr2
ǫo. (b) Find the total charge contained in the sphere.
Soln. :
The total charge is obtained by integrating the charge density
R
0
ρdτ = 5ǫok r2
4πr2
dr = 4πkǫ0R5
12
The same result is also obtained by the surface integral of the electric field :
E · dS = kR3
ˆr · (R2
sin θdθdφˆr) = 4πkR5
=
Q
ǫ0
The two results are consistent.
Example 2 :
Calculate the flux through the shaded
area (face of a cube of side a) when a
charge q is located at one of the dis-
tant corners from the side.
000000000000
00000000
0000
00000000
0000
00000000
00000000
0000
00000000
00000000
111111111111
11111111
1111
11111111
1111
11111111
11111111
1111
11111111
11111111
q
Solution :
If the charge were located at the centre of the cube instead of the corner, the flux
would have been q/6ǫ0, by symmetry. To use this symmetry consider the given
cube to be a part of a bigger cube of side 2a × 2a, as shown, so that the charge q
is in the centre of the bigger cube.
The flux through each face of the
begger cube is now q/6ǫ0. Be-
cause the side of the bigger cube
consists of four identical faces, the
flux through one fourth of the face is
clearly q/24ǫ0.
000
0000
0
111
1111
1
Applications of Gauss’s Law
Field due to a uniformly charged sphere of radius R with a charge Q
By symmetry, the field is radial. Gaussian surface is a concentric sphere of radius
r. The outward normals to the Gaussian surface is parallel to the field E at every
point. Hence E · dS = 4πr2
| E |
13
Q
r
R
E
R
rQ
E
r >R r < R
For r > R,
4πr2
| E |=
Q
r2
so that
E =
Q
4πr2
ˆr
The field outside the sphere is what it would be if all the charge is concentrated at
the origin of the sphere.
14
For r < R, a fraction r3
/R3
of the
total charge is enclosed within the
gaussian surface, so that
4πr2
| E |=
1
ǫ0
Qr3
R3
The field inside is
E =
Q
4πǫ0
r
R3
ˆr
Einarbitraryunits
r--->
Field due to sphere with charge density ρ
R
Exercise :
Find the electric field both inside and outside a spherical shell of radius R carrying
a uniform charge Q.
Example 3 :
Find the electric field inside a sphere of radius R which carries a charge density
ρ = kr where r is the distance from the origin and k is a constant.
Solution :
By symmetry the field is radial. Take the gaussian surface to be a sphere of radius
R. The flus is 4πr2
| E |.
15
r
R
The charge enclosed by the gaussian surface is
Q =
r
0
ρ(r)d3
r =
r
0
ρ(r)4πr2
dr
= 4πk
r
0
r3
dr
= πkr4
Thus
E =
1
ǫ0
kr2
4
ˆr
(what is the dimension of k ?)
Exercise : A hollow spherical shell carries a charge density ρ = k/r2
for
a ≤ r ≤ b. Calculate the field at all points. (Ans. For r < a, field is zero, for
a < r < b, | E |= k(r − a)/ǫ0r2
, and for r > b, | E |= k(b − a)ǫ0r2
.)
Field due to an infinite line charge of linear charge density λ
Gaussian surface is a cylinder of radius r and length L.
By symmetry, the field has the same magnitude at every point on the curved sur-
face and is directed outwards. At the end caps, E is perpendicular to dS every-
where and the flux is zero. For the curved surface, E and dS are parallel,
16
dS1
dS2
E
L
++++++++++++++++
E · dS = | E | .2πrL
=
Q
ǫ0
=
λL
ǫ0
Thus
E =
λ
2πǫ0r
ˆρ
where ˆρ is a unit vector perpendicular to the line,directed outward for positive line
charge and inward for negative line charge.
Exercise :
Find the electric field both inside and outside a long cylinder of radius R carrying
a uniform volume charge density ρ.
(Hint : Take the gaussian surface to be a finite concentric cylinder of radius r (with
r < R and r > R), as shown)
+ + + + + + +++ +
3 Field due to an infinite charged sheet with surface charge density σ
Choose a cylindrical Gaussian pillbox of height h (with h/2 above the sheet and
h/2 below the sheet) and radius r.
17
dS
E
+
+
+
++
+
+ + +
+
+
+
++ +
+
+
+
+ +
+
++++
+
L
σ
dSE
E
r
dS
The amount of charge enclosed is area times the surface charge density, i.e., Q =
πr2
σ. By symmetry, the field is directed perpendicular to the sheet, upward at
points above the sheet and downward for points below. There is no contribution
to the flux from the curved surface. The flux from the two end faces is πr2
| E |
each, i.e. a total outward flux of 2πr2
| E |. Hence
2πr2
| E |=
Q
ǫ0
=
πr2
σ
ǫ0
so that
E =
σ
2ǫ0
ˆn
where ˆn is a unit vector perpendicular to the sheet, directed upward for points
above and downwards for points below (opposite, if the charge density is nega-
tive).
Exercise :
Find the electric field in the region between two infinite parallel planes carrying
charge densities +σ and −σ.
18
+
+
+
+
+
+
+
+
+σ −σ
Exercise :
A very long cylinder carries a charge density ρ = kr, where r is the distance from
the axis of the cylinder. Find the electric field at a distance r < R. (Ans.
(1/3ǫ0)kr2
ˆr)
Example 4 :
Two spheres of radius R each overlap such that the distance between their centres
is separated by a distance 2R − s. Show that the field in the overlapping region is
constant.
Solution :
O O’
s
θ
φ
P
Q
The figure shows the field at a point P in the overlap region due to the two spheres.
Taking the expression for field at a point inside the sphere and resolving into x-
and y- components, The x- components reinforce while the y-components are in
19
the opposite directions. It can be seen,
Ey =
Q
4πǫ0
1
R3
(OP sin θ − O′
P sin φ)
Using the property of triangles (the ratio of the sides is equal to the ratio of the
sine of opposite angles a/ sin A = b/ sin B) the y-component is seen to vanish.
The x-component of the field is
Ex =
Q
4πǫ0
1
R3
(OP cos θ + O′
P cos φ)
=
Q
4πǫ0
1
R3
(OQ + O′
Q)
=
Q
4πǫ0
1
R3
s
Thus the field depends only on the distance between the centres and is constant.
Example 5 :
A sphere of radius R has a cavity of radius a inside it. The sphere has uniform
charge density ρ spread over its volume. Show that the field inside the cavity is
constant.
Solution :
One can use superposition princi-
ple to solve this problem. Fill up
the cavity with equal and opposite
charge distribution. The problenm
then is equivalent to the field due to
a sphere of radius R and charge den-
sity ρ and a smaller sphere of radius
a, but with a charge densiuty −ρ.
R
a
d
r
O
P
O’
We calculate the fields due to these two spheres at r (with respect to O, the centre
20
of the larger sphere, The field due to larger sphere
E1 =
Q
4πǫ0
r
R3
=
ρ
4πǫ0
4π
3
R2 r
R3
=
ρ
3ǫ0
r
By identical argument, the field due to smaller sphere (the point P is at r − d with
respect to the centre of the smaller sphere),
E2 = −
ρ
3ǫ0
(r − d)
Adding,
E =
ρ
3ǫ0
d = constant
21

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Electrostatic

  • 1. ELECTROSTATICS Gauss’s Law and Applications Though Coulomb’s law is fundamental, one finds it cumbersome to use it to cal- culate electric field due to a continuous charge distribution because the integrals involved can be quite difficult. An alternative but completely equivalent formula- tion is Gauss’s Law which is very useful in situations which exhibit certain sym- metry. Electric Lines of Force : Electric lines of force (also known as field lines) is a pictorial representation of the electric field. These consist of directed lines indicating the direction of electric field at various points in space. • There is no rule as to how many lines are to be shown. However, it is customary to draw number of lines proportional to the charge. Thus if N number of lines are drawn from or into a charge Q, 2N number of lines would be drawn for charge 2Q. • The electric field at a point is directed along the tangent to the field lines. A positive charge at this point will move along the tangent in a direction indicated by the arrow. • Lines are dense close to a source of the electric field and become sparse as one moves away. • Lines originate from a positive charge and end either on a negative charge or move to infinity. 1
  • 2. • Lines of force due to a solitary negative charge is assumed to start at infinity and end at the negative charge. • Field lines do not cross each other. ( if they did, the field at the point of crossing will not be uniquely defined.) • A neutral point is a point at which field strength is zero. This occurs because of cancellation of electric field at such a point due to multiple charges. Exercise : Draw field lines and show the neutral point for a charge +4Q located at (1, 0) and −Q located at (−1, 0). 2.3 Electric Flux The concept of flux is borrowed from flow of water through a surface. The amount of water flowing through a surface depends on the velocity of water, the area of the surface and the orientation of the surface with respect to the direction of velocity of water. Though an area is generally considered as a scalar, an element of area may be considered to be a vector because : • It has magnitude (measured in m2 ). • If the area is infinitisimally small, it can be considered to be in a plane. We can then associate a direction with it. For instance, if the area element lies in the x-y plane, it can be considered to be directed along the z–direction. (Conventionally, the direction of the area is taken to be along the outward normal.) 2
  • 3. S θ In the figure above, the length of the vector S is chosen to represent the area in some convenient unit and its direction is taken to be along the outward normal to the area. We define the flux of the electric field through an area dS to be given by the scalar product dφ = E · dS If θ is the angle between the electric field and the area vector dφ =| E || dS | cos θ . For an arbitrary surface S, the flux is obtainted by integrating over all the surface elements φ = dφ = S E · dS If the electric field is uniform, the angle θ is constant and we have φ = ES cos θ = E(S cos θ) Thus the flux is equal to the product of magnitude of the electric field and the projection of area perpendicular to the field. 3
  • 4. S E θ Unit of flux is N-m2 /C. Flux is positive if the field lines come out of the surface and is negative if they go into it. Solid Angle : The concept of solid angle is a natural extension of a plane angle to three dimen- sions. Consider an area element dS at a distance r from a point P. Let ˆn be the unit vector along the outward normal to dS. The element of the solid angle sub- tended by the area element at P is de- fined as dΩ = dS⊥ r2 where dS⊥ is the projection of dS along a direction perpendicular to r. If α is the angle between ˆr and ˆn, then, dΩ = dS cos α r2 r n^ ^ P d Ω α dS Solid angle is dimensionless. However, for practical reasons it is measured in terms of a unit called steradian (much like the way a planar angle is measured in terms of degrees). The maximum possible value of solid angle is 4π, which is the angle subtended by an area which encloses the point P completely. Example 1: A right circular cone has a semi-vertical angle α. Calculate the solid angle at the apex P of the cone. Solution : The cap on the cone is a part of a sphere of radius R, the slant length of the cone. Using spherical polar coordinates, an area element on the cap is R2 sin θdθdφ, where θ is the polar angle and φ is the azimuthal angle. Here, φ goes from 0 to 2π 4
  • 5. while θ goes from 0 to α. Thus the area of the cap is dA = 2πR2 α 0 sin θdθ = 2πR2 (1 − cos α) Thus the solid angle at P is dΩ = dA R2 = 2π(1 − cos α) R P θ Exercise : Calculate the solid angle subtended by an octant of a sphere at the centre of the sphere. (Ans. π/2) The flux per unit solid angle is known as the intensity. Example 2 An wedge in the shape of a rectangular box is kept on a horizontal floor. The two triangular faces and the rectangular face ABFE are in the vertical plane. The electric field is horizontal, has a magnitude 8 × 104 N/C and enters the wedge through the face ABFE, as shown. Calculate the flux through each of the faces and through the entire surface of the wedge. 000000000000000000000000 000000000000000000000000 000000000000000000000000000000000000 111111111111111111111111 111111111111111111111111 111111111111111111111111111111111111 0.4 m 0.3m 0.2m E A B C DE F Solution : The outward normals to the triangular faces AED, BFC, as well as the normal to the base are perpendicular to E. Hence the flux through each of these faces is zero. The vertical rectangular face ABFE has an area 0.06 m2 . The outward normal to this face is perpendicular to the electric field. The flux is entering through this 5
  • 6. face and is negative. Thus flux through ABFE is φ1 = −0.06 × 8 × 104 = −.48 × 104 N − m2 /C To find the flux through the slanted face, we need the angle that the normal to this face makes with the horizontal electric field. Since the electric field is perpen- dicular to the side ABFE, this angle is equal to the angle between AE and AD, which is cos−1 (.3/.5). The area of the slanted face ABCD is 0.1 m2 . Thus the flux through ABCD is φ2 = 0.1 × 8 × 104 × (.3/.5) = +0.48 × 104 N − m2 /C The flux through the entire surface of the wedge is φ1 + φ2 = 0. Example 3: Calculate the flux through the base of the cone of radius R. E Solution : The flux entering is perpendicular to the base. Since the outward normal to the cir- cular base is in the opposite sense, the flux is negative and is equal to the product of the magnitude of the field and the area of the base, The flux, therefore is, πR2 E. Example 4 : Calculate the flux coming out through the curved surface of the cone in the above example. Solution : 6
  • 7. dl h H r θ R Consider a circular strip of radius r at a depth h from the apex of the cone. The angle between the electric field through the strip and the vector dS is π −θ, where θ is the semi-angle of the cone. If dl is the length element along the slope, the area of the strip is 2πrdl. Thus, E · dS = 2πrdl | E | sin θ We have, l = h/ cos θ, so that dl = dh/ cos θ. Further, r = h tan θ Substituting, we get E · dS = 2πh tan2 θ | E | dh Integrating from h = 0 to h = H, the height of the cone, the outward flux is | E | H2 tan2 θ = πR2 | E |. Example 5 : A charge Q is located at the center of a sphere of radius R. Calculate the flux going out through the surface of the sphere. dS By Coulomb’s law, the field due to the charge Q is radial and is given on the surface of the sphere by, E = 1 4πǫ0 Q R2 ˆr 7
  • 8. The direction of the area vector dS, is also radial at each point of the surface dS = dSˆr. The flux φ = E · dS = 1 4πǫ0 Q R2 dS The integral over dS is equal to the surface area of the sphere, which is, 4πR2 . Thus the flux out of the surface of the sphere is φ = 1 4πǫ0 Q R2 · 4πR2 = Q ǫ0 Note that the flux is independent of the radius of the sphere - a cancellation due to the fact that the surface area of the sphere is proportional to r2 while the field is proportional to 1/r2 . Curiously, the result is valid for any arbitrar- ily shaped surface. Consider a cone of solid angle dΩ centered at the charge. This will intersect the arbitrarily shaped surface in an ellipse whose normal ˆn makes an angle θ with the outgoing radial direction ˆr. If the area of the ellipse is dS, dΩ = dS cos θ r2 so that the flux through the cone is dΦ = 1 4πǫ0 q r2 ˆr · dS = 1 4πǫ0 q r2 ˆr · r2 dΩ cos θ ˆn = 1 4πǫ0 qdΩ ˆn · r cos θ = qdΩ 4πǫ0 (1) q dS r n^ ^ r dΩ Thus the total flux throgh the surface is Φ = dΦ = q 4πǫ0 dΩ = q ǫ0 (2) 8
  • 9. One can generalize this to multiple charges since superposition principle holds for the electric field. Suppose the total field consists of fields E1 due to charge q1, E2 due to q2 and so on. We have E · dS = i Ei · dS = i qi ǫ0 One should note that the above derivation is valid only if the charges are contained within the volume for only then the total solid angle be- comes 4π. For charges which are outside the volume, the flux that en- ters the volume also leaves it and though the areas which the cone in- tersects are different, the solid angles are the same. This leads to cancella- tion of the flux and the contribution to the flux from a charge which re- sides outside the volume is zero. r dΩ q r^ n^ n^ dS dS1 1 2 2 GAUSS’S LAW - Integral form The flux calculation done in Example 4 above is a general result for flux out of any closed surface, known as Gauss’s law. Total outward electric flux φ through a closed surface S is equal to 1/ǫ0 times the charge enclosed by the volume defined by the surface S 9
  • 10. 01E 01E 01E 00001111 01dS Mathematicaly, the surface integral of the electric field over any closed surface is equal to the net charge enclosed divided by ǫ0 E · dS = Qenclosed ǫ0 (3) • The law is valid for arbitry shaped surface, real or imaginary. • Its physical content is the same as that of Coulomb’s law. • In practice, it allows evaluation of electric field in many practical situa- tions by forming imagined surfaces which exploit symmetry of the problem. Such surfaces are called Gaussian surfaces. GAUSS’S LAW - Differential form The integral form of Gauss’s law can be converted to a differential form by using the divergence theorem. If V is the volume enclosed by the surface S, S E · dS = V ∇ · Edv (A) If ρ is the volume charge density, Q = V ρdv (B) 10
  • 11. Thus we have ∇ · E = ρ ǫ0 (4) Direct Calculation of divergence from Coulomb’s Law : We will use the field expression E(r) = 1 4πǫ0 (r − r′ ) | r − r′ |3 ρ(r′ )d3 r to directly evaluate the divergence of the electric field. Since the differentiation is with respect to r while the integration is with respect to r′ , we can take the divergence inside the integral, ∇ · E(r) = 1 4πǫ0 ∇r · (r − r′ ) | r − r′ |3 ρ(r′ )d3 r A simple minded calculation of divergence is as follows : ∇r · (r − r′ ) | r − r′ |3 = (∇r · r) 1 | r − r′ |3 + (r − r′ ) · ∇r 1 | r − r′ |3 Divergence of r is equal to 3 ∇r · r = ∂ ∂x x + ∂ ∂y y + ∂ ∂z z = 3 ∇r 1 | r − r′ |3 = −3 r − r′ | r − r′ |5 Thus it would seem that ∇r · (r − r′ ) | r − r′ |3 = 0 i.e. ∇ · E = 0, which violates Gauss’s Law, that ∇ · E = ρ/ǫ0. The problem arises because the function 1/ | r − r′ | has a singularity at r = r′ . This point has to be taken care of with care. Except at this point the divergence of the integral is indeed zero. 11
  • 12. Thus we can shrink the range of in- tegral till it becomes a small sphere around the point r(x, y, z). Since ρ(x′ , y′ , z′ ) is continuous, we may replace the density on the surface of the small sphere by the value of den- sity at the centre, so that the density term can come out of the integral, leaving, ∇·E(r) = 1 4πǫ0 ρ(r) ∇r· (r − r′ ) | r − r′ |3 d3 r r − r’ (x,y,z) (x’,y’,z’) Since the divergence with respect to r is being taken of a function which only depends on the difference r − r′ , we may replace ∇r → −∇′ r, which gives ∇ · E(r) = − 1 4πǫ0 ρ(r) ∇r′ · (r − r′ ) | r − r′ |3 d3 r The volume integral on the right may be converted to a surface integral using the divergence theore, ∇ · E(r) = − 1 4πǫ0 ρ(r) (r − r′ ) | r − r′ |3 · dS where the integral is over the surface of the sphere. If the radius of the sphere be taken as r0, r − r′ = −r0ˆn and dS = dSˆn, Hence ∇ · E = + 1 4πǫ0 ρ(r) dS r2 0 = 1 4πǫ0 ρ(r) · 4πr2 0 = ρ(r) ǫ0 Example 1: A sphere of radius R contains a continuous charge distribution ρ(r). The electric field at a distance r from the centre of the sphere is E = kr3 ˆr. (a) Find the charge density. Soln. : ∇E = 1 r2 ∂ ∂r (r2 Er) = 1 r2 ∂ ∂r (kr5 ) = 5kr2 = ρ/ǫo Thus ρ = 5kr2 ǫo. (b) Find the total charge contained in the sphere. Soln. : The total charge is obtained by integrating the charge density R 0 ρdτ = 5ǫok r2 4πr2 dr = 4πkǫ0R5 12
  • 13. The same result is also obtained by the surface integral of the electric field : E · dS = kR3 ˆr · (R2 sin θdθdφˆr) = 4πkR5 = Q ǫ0 The two results are consistent. Example 2 : Calculate the flux through the shaded area (face of a cube of side a) when a charge q is located at one of the dis- tant corners from the side. 000000000000 00000000 0000 00000000 0000 00000000 00000000 0000 00000000 00000000 111111111111 11111111 1111 11111111 1111 11111111 11111111 1111 11111111 11111111 q Solution : If the charge were located at the centre of the cube instead of the corner, the flux would have been q/6ǫ0, by symmetry. To use this symmetry consider the given cube to be a part of a bigger cube of side 2a × 2a, as shown, so that the charge q is in the centre of the bigger cube. The flux through each face of the begger cube is now q/6ǫ0. Be- cause the side of the bigger cube consists of four identical faces, the flux through one fourth of the face is clearly q/24ǫ0. 000 0000 0 111 1111 1 Applications of Gauss’s Law Field due to a uniformly charged sphere of radius R with a charge Q By symmetry, the field is radial. Gaussian surface is a concentric sphere of radius r. The outward normals to the Gaussian surface is parallel to the field E at every point. Hence E · dS = 4πr2 | E | 13
  • 14. Q r R E R rQ E r >R r < R For r > R, 4πr2 | E |= Q r2 so that E = Q 4πr2 ˆr The field outside the sphere is what it would be if all the charge is concentrated at the origin of the sphere. 14
  • 15. For r < R, a fraction r3 /R3 of the total charge is enclosed within the gaussian surface, so that 4πr2 | E |= 1 ǫ0 Qr3 R3 The field inside is E = Q 4πǫ0 r R3 ˆr Einarbitraryunits r---> Field due to sphere with charge density ρ R Exercise : Find the electric field both inside and outside a spherical shell of radius R carrying a uniform charge Q. Example 3 : Find the electric field inside a sphere of radius R which carries a charge density ρ = kr where r is the distance from the origin and k is a constant. Solution : By symmetry the field is radial. Take the gaussian surface to be a sphere of radius R. The flus is 4πr2 | E |. 15
  • 16. r R The charge enclosed by the gaussian surface is Q = r 0 ρ(r)d3 r = r 0 ρ(r)4πr2 dr = 4πk r 0 r3 dr = πkr4 Thus E = 1 ǫ0 kr2 4 ˆr (what is the dimension of k ?) Exercise : A hollow spherical shell carries a charge density ρ = k/r2 for a ≤ r ≤ b. Calculate the field at all points. (Ans. For r < a, field is zero, for a < r < b, | E |= k(r − a)/ǫ0r2 , and for r > b, | E |= k(b − a)ǫ0r2 .) Field due to an infinite line charge of linear charge density λ Gaussian surface is a cylinder of radius r and length L. By symmetry, the field has the same magnitude at every point on the curved sur- face and is directed outwards. At the end caps, E is perpendicular to dS every- where and the flux is zero. For the curved surface, E and dS are parallel, 16
  • 17. dS1 dS2 E L ++++++++++++++++ E · dS = | E | .2πrL = Q ǫ0 = λL ǫ0 Thus E = λ 2πǫ0r ˆρ where ˆρ is a unit vector perpendicular to the line,directed outward for positive line charge and inward for negative line charge. Exercise : Find the electric field both inside and outside a long cylinder of radius R carrying a uniform volume charge density ρ. (Hint : Take the gaussian surface to be a finite concentric cylinder of radius r (with r < R and r > R), as shown) + + + + + + +++ + 3 Field due to an infinite charged sheet with surface charge density σ Choose a cylindrical Gaussian pillbox of height h (with h/2 above the sheet and h/2 below the sheet) and radius r. 17
  • 18. dS E + + + ++ + + + + + + + ++ + + + + + + + ++++ + L σ dSE E r dS The amount of charge enclosed is area times the surface charge density, i.e., Q = πr2 σ. By symmetry, the field is directed perpendicular to the sheet, upward at points above the sheet and downward for points below. There is no contribution to the flux from the curved surface. The flux from the two end faces is πr2 | E | each, i.e. a total outward flux of 2πr2 | E |. Hence 2πr2 | E |= Q ǫ0 = πr2 σ ǫ0 so that E = σ 2ǫ0 ˆn where ˆn is a unit vector perpendicular to the sheet, directed upward for points above and downwards for points below (opposite, if the charge density is nega- tive). Exercise : Find the electric field in the region between two infinite parallel planes carrying charge densities +σ and −σ. 18
  • 19. + + + + + + + + +σ −σ Exercise : A very long cylinder carries a charge density ρ = kr, where r is the distance from the axis of the cylinder. Find the electric field at a distance r < R. (Ans. (1/3ǫ0)kr2 ˆr) Example 4 : Two spheres of radius R each overlap such that the distance between their centres is separated by a distance 2R − s. Show that the field in the overlapping region is constant. Solution : O O’ s θ φ P Q The figure shows the field at a point P in the overlap region due to the two spheres. Taking the expression for field at a point inside the sphere and resolving into x- and y- components, The x- components reinforce while the y-components are in 19
  • 20. the opposite directions. It can be seen, Ey = Q 4πǫ0 1 R3 (OP sin θ − O′ P sin φ) Using the property of triangles (the ratio of the sides is equal to the ratio of the sine of opposite angles a/ sin A = b/ sin B) the y-component is seen to vanish. The x-component of the field is Ex = Q 4πǫ0 1 R3 (OP cos θ + O′ P cos φ) = Q 4πǫ0 1 R3 (OQ + O′ Q) = Q 4πǫ0 1 R3 s Thus the field depends only on the distance between the centres and is constant. Example 5 : A sphere of radius R has a cavity of radius a inside it. The sphere has uniform charge density ρ spread over its volume. Show that the field inside the cavity is constant. Solution : One can use superposition princi- ple to solve this problem. Fill up the cavity with equal and opposite charge distribution. The problenm then is equivalent to the field due to a sphere of radius R and charge den- sity ρ and a smaller sphere of radius a, but with a charge densiuty −ρ. R a d r O P O’ We calculate the fields due to these two spheres at r (with respect to O, the centre 20
  • 21. of the larger sphere, The field due to larger sphere E1 = Q 4πǫ0 r R3 = ρ 4πǫ0 4π 3 R2 r R3 = ρ 3ǫ0 r By identical argument, the field due to smaller sphere (the point P is at r − d with respect to the centre of the smaller sphere), E2 = − ρ 3ǫ0 (r − d) Adding, E = ρ 3ǫ0 d = constant 21