3. A positive point charge Q is surrounded by an
imaginary sphere of radius r
The total electric flux through the closed surface of
the sphere using the equation
The electric field of the point charge is directed
radially outward at all points on the surface of the
sphere
Therefore, the direction of the area element dA is
along the electric field E and .θ= 00
4. E is uniform on the surface of the sphere
The equation is called as Gauss’s law
6. It is seen that the total electric flux is the same for
closed surfaces A1, A2 and A3
Gauss’s law states that if a charge Q is enclosed
by an arbitrary closed surface, then the total
electric flux ΦE through the closed surface is
Here Qencl denotes the charges inside the closed
surface
7. Discussion of Gauss law
The total electric flux through the closed surface
depends only on the charges enclosed by the
surface and the charges present outside the surface
will not contribute to the flux and the shape of the
closed surface which can be chosen arbitrarily
The total electric flux is independent of the location
of the charges inside the closed surface
8. This imaginary surface is called a Gaussian surface.
The shape of the Gaussian surface to be chosen
depends on the type of charge configuration and the
kind of symmetry existing in that charge
configuration. The electric field is spherically
symmetric for a point charge, therefore spherical
Gaussian surface is chosen. Cylindrical and planar
Gaussian surfaces can be chosen for other kinds of
charge configurations
9. The electric field is due to charges present inside
and outside the Gaussian surface but the charge
Qencl denotes the charges which lie only inside the
Gaussian surface
The Gaussian surface cannot pass through any
discrete charge but it can pass through continuous
charge distributions. It is because, very close to the
discrete charges, the electric field is not well defined
Gauss law is another form of Coulomb’s law and it is
also applicable to the charges in motion. Because of
this reason, Gauss law is treated as much more
general law than Coulomb’s law.
10.
11.
12.
13. Electric field due to an infinitely
long charged wire
Consider an infinitely long straight wire having
uniform linear charge density λ.
Let P be a point located at a perpendicular distance r
from the wire
The electric field at the point P can be found using
Gauss law.
We choose two small charge elements A1 and A2 on
the wire which are at equal distances from the point
P.
The resultant electric field due to these two charge
elements points radially away from the charged wire
and the magnitude of electric field is same at all
points on the circle of radius r.
17. Let us choose a cylindrical Gaussian surface of radius r
and length L
The total electric flux in this closed surface is
calculated as follows.
18. Since the magnitude of the electric field for the entire
curved surface is constant, E is taken out of the
integration and Qencl and is given by Qencl =λL
19. The electric field is always along the perpendicular
direction (r) to wire.
20. The equation is true only for an infinitely long
charged wire.
For a charged wire of finite length, the electric field
need not be radial at all points.
However, equation for such a wire is taken
approximately true around the mid-point of the wire
and far away from the both ends of the wire
21. Electric field due to charged
infinite plane sheet
Consider an infinite plane sheet of charges with
uniform surface charge density σ.
Let P be a point at a distance of r from the sheet
Since the plane is infinitely large, the electric field
should be same at all points equidistant from the
plane and radially directed at all points.
A cylindrical shaped Gaussian surface of length 2r
and area A of the flat surfaces is chosen such that
the infinite plane sheet passes perpendicularly
through the middle part of the Gaussian surface
23. Applying Gauss law for this cylindrical surface ′
The electric field is perpendicular to the area
element at all points on the curved surface and is
parallel to the surface areas at P andP’
24. Since the magnitude of the electric field at these two
equal surfaces is uniform, E is taken out of the
integration andQencl is given by , Qencl=σ A
25. The electric field due to an infinite plane
sheet of charge depends on the surface
charge density and is independent of the
distance r.
The electric field will be the same at any
point farther away from the charged plane.
If σ > 0 the electric field at any point P is
outward perpendicular to the plane
If σ < 0 the electric field points inward
perpendicularly (-n) to the plane.
26. Electric field due to two parallel
charged infinite sheets
Consider two infinitely large charged plane sheets
with equal and opposite charge densities +σ and -σ
which are placed parallel to each other
The electric field between the plates and outside the
plates is found using Gauss law.
The magnitude of the electric field due to an infinite
charged plane sheet is σ /2ε0 and it points
perpendicularly outward if σ > 0 and points inward if
σ < 0.
28. At the points P2 and P3, the electric field due to both
plates are equal in magnitude and opposite in
direction .
As a result, electric field at a point outside the plates
is zero. But inside the plate, electric fields are in
same direction i.e., towards the right, the total
electric field at a point P1
The direction of the electric field inside the plates is
directed from positively charged plate to negatively
charged plate and is uniform everywhere inside the
plate
29. Electric field due to a uniformly
charged spherical shell
Consider a uniformly charged spherical shell of
radius R and total charge Q
The electric field at points outside and inside the
sphere is found using Gauss law.
Case (a) At a point outside the shell (r > R)
Let us choose a point P outside the shell at a
distance r from the center
The charge is uniformly distributed on the surface of
the sphere (spherical symmetry).
Hence the electric field must point radially outward if
Q > 0 and point radially inward if Q < 0.
30.
31. So we choose a spherical Gaussian surface of
radius r is chosen and the total charge enclosed by
this Gaussian surface is Q. Applying Gauss law
32.
33. (b) At p i t on he sur ace of he sph rical she l (r R
he
34.
35. The electric field due to the uniformly charged
spherical shell is zero at all points inside the shell
A graph is plotted between the electric field and radial
distance
