APPLICATION OF GAUSS’S LAW

Now that we are aware of Gauss’s law, it is time to find some usefulness of the law. This law helps us in
knowing the electric field intensity due to various kinds of charge distributions. Let us see how.

Following steps will form the general procedure for determining the unknown electric field intensities.

1. Choose a Gaussian surface, that is, choose a closed surface that will be considered for evaluating LHS
and RHS of Gauss’s law.

2. Evaluate LHS.

3. Evaluate RHS.

4. Equate LHS to RHS.

As a start we will consider a case for which we already know the expression of electric field intensity.
This will help us build the trust in Gauss’s law.

1. Electric field intensity due to a Charged particle.

Our aim is to derive electric field strength at point P that lies at a distance r from particle.

Step 1. We will choose a spherical Gaussian surface with centre at charge q and radius r. Such a surface
will pass through P.

Advantages of this surface:

(a) As all points of this surface are equidistant from the charged particle, therefore by symmetry we can
assert that the magnitude of electric field strength at all these points must be same (which is unknown
to us right now).
(b) At all points of this surface, the local area vector is parallel to electric field direction. Thus θ is zero at
all points of surface.

Step 2. LHS = ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∮ 𝐸 𝑑𝐴 cos 0 = 𝐸 ∮ 𝑑𝐴 = 𝐸 𝑋 4𝜋𝑟 2

𝑞𝑒𝑛 𝑞
Step 3. RHS = =
𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

𝑞
⇒ 𝐸 𝑋 4𝜋𝑟2 =
𝜖𝑜
𝑞
⇒𝐸= = 𝑘𝑞/𝑟 2
4𝜋𝑟 2 𝜖𝑜

This is the same good old expression that we are already aware of! Gauss’s law works!

2. Electric field strength due to a uniform distribution of charge on a spherical surface.

Figure shows the case under consideration. We will divide this case into two parts- (a) Points
that lie outside the spherical surface and (b) points that lie inside the spherical surface.

Case (a) At a point P outside the spherical surface

Step 1 . We will choose a concentric spherical Gaussian surface of radius r. Such a surface will pass
through P.

Advantages of this surface:

(a) As all points of this surface are equidistant from the centre, therefore by symmetry we can assert
that the magnitude of electric field strength at all these points must be same (which is unknown to us
right now).

(b) At all points of this surface, the local area vector is parallel to electric field direction. Thus θ is zero at
all points of surface.

⃗⃗⃗⃗⃗ = ∮ 𝐸 𝑑𝐴 cos 0 = 𝐸 ∮ 𝑑𝐴 = 𝐸 𝑋 4𝜋𝑟 2

Step 2. LHS = ∮ 𝐸⃗ . 𝑑𝐴

𝑞𝑒𝑛 𝑞
Step 3. RHS = = , here ‘q’ represents the total charge present on the spherical surface.
𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

𝑞
⇒ 𝐸 𝑋 4𝜋𝑟2 =
𝜖𝑜
𝑞
⇒𝐸= = 𝑘𝑞/𝑟 2
4𝜋𝑟 2 𝜖𝑜

This expression is same as that of a charged particle.

Case (b) At a point P inside the spherical surface

Step 1 . We will choose a concentric spherical Gaussian surface of radius r. Such a surface will pass
through P.

Advantages of this surface:

(a) As all points of this surface are equidistant from the centre, therefore by symmetry we can assert
that the magnitude of electric field strength at all these points must be same (which is unknown to us
right now).

(b) At all points of this surface, the local area vector is parallel to electric field direction. Thus θ is zero at
all points of surface.

⃗⃗⃗⃗⃗ = ∮ 𝐸 𝑑𝐴 cos 0 = 𝐸 ∮ 𝑑𝐴 = 𝐸 𝑋 4𝜋𝑟 2

Step 2. LHS = ∮ 𝐸⃗ . 𝑑𝐴

𝑞𝑒𝑛 0
Step 3. RHS = =
𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

0
⇒ 𝐸 𝑋 4𝜋𝑟2 =
𝜖𝑜
⇒𝐸= 0

Thus electric field strength inside a uniform distribution of charge on a spherical surface due to
itself is zero.

𝑘𝑞
𝐸= ,𝑟 ≥ 𝑅
𝑟2
0, 𝑟 <𝑅

Both the cases can be represented in a E vs r graph as shown. ( “R” represents the radius of the
surface)
3. Electric field strength due to a uniform distribution of charge in a spherical volume.

This time the charge is distributed throughout the volume of the sphere. In such a case, we may
be given the radius ‘R’ and the totals charge ‘Q’ or the volume charge density ‘ρ’. The three are
related to one another as

𝑄
𝜌= 4
𝜋𝑅 3
3

So if any two out of the three (’R’, ‘Q’ and ‘ρ’) is given then the third can be evaluated by using
above relation. Usual practice is to mention radius ’R’ along with either of the remaining two.

For this case also we will consider two parts- (a) Points that lie outside the spherical region and
(b) points that lie inside the spherical region.

Case (a) At a point P outside the spherical region

Step 1 . We will choose a concentric spherical Gaussian surface of radius r. Such a surface will pass
through P.

Advantages of this surface:

(a) As all points of this surface are equidistant from the centre, therefore by symmetry we can assert
that the magnitude of electric field strength at all these points must be same (which is unknown to us
right now).

(b) At all points of this surface, the local area vector is parallel to electric field direction. Thus θ is zero at
all points of surface.

Step 2. LHS = ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∮ 𝐸 𝑑𝐴 cos 0 = 𝐸 ∮ 𝑑𝐴 = 𝐸 𝑋 4𝜋𝑟 2

𝑞𝑒𝑛 𝑄
Step 3. RHS = = , here ‘Q’ represents the total charge present in the spherical region.
𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

𝑄
⇒ 𝐸 𝑋 4𝜋𝑟2 =
𝜖𝑜
𝑄
⇒𝐸= = 𝑘𝑄/𝑟 2
4𝜋𝑟 2 𝜖𝑜

This expression is same as that of a charged particle.

Case (b) At a point P inside the spherical region

Step 1 . We will choose a concentric spherical Gaussian surface of radius r. Such a surface will pass
through P.

Advantages of this surface:

(a) As all points of this surface are equidistant from the centre, therefore by symmetry we can assert
that the magnitude of electric field strength at all these points must be same (which is unknown to us
right now).

(b) At all points of this surface, the local area vector is parallel to electric field direction. Thus θ is zero at
all points of surface.

⃗⃗⃗⃗⃗ = ∮ 𝐸 𝑑𝐴 cos 0 = 𝐸 ∮ 𝑑𝐴 = 𝐸 𝑋 4𝜋𝑟 2

Step 2. LHS = ∮ 𝐸⃗ . 𝑑𝐴
4
𝑞𝑒𝑛 𝜌 𝑋 3𝜋𝑟3
Step 3. RHS = =
𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

𝜌 𝑋 43 𝜋𝑟 3
⇒ 𝐸 𝑋 4𝜋𝑟2 =
𝜖𝑜
𝜌𝑟 𝑘𝑄𝑟
⇒𝐸= = 3
3𝜖𝑜 𝑅

Thus the expression for E would be

𝑘𝑄
𝐸= ,𝑟 ≥ 𝑅
𝑟2
𝜌𝑟 𝑘𝑄𝑟
= 3 ,𝑟 < 𝑅
3𝜖𝑜 𝑅
Both the cases can be represented in an E vs r graph as shown.
NOTE: The electric field inside the spherical region can be written in vector form as
𝜌𝑟
𝐸⃗𝑖𝑛𝑛𝑒𝑟 =
3𝜖𝑜

Here ρ is to be substituted with sign.

4. Electric field strength due to a uniformly charged wire of infinite length

This is again a case for which the expression is already known to us. But still, we will see the
derivation of expression by means of Gauss’s law here. Earlier the result was derived using
calculus.

The electric field pattern for this case would be somewhat as depicted in figure. The field
pattern can be judged on basis of symmetry.
Step 1. We will choose a cylindrical Gaussian surface. This cylindrical surface is co-axial to given
wire and has a radius ‘r’. Thus this surface will pass through P. Note that the chosen surface
should be a closed surface. So we will also assume two discs of radius r that will serve as lid and
base of our cylindrical surface. The height of cylindrical surface is denoted as ‘h’. So the
cylindrical Gaussian surface would look as shown in figure.

Step 2. LHS = ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∫𝐷𝑖𝑠𝑐1 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝐷𝑖𝑠𝑐2 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴

In above step, we have divided the closed surface into three open surfaces. First one is upper
disc (lid) named as “Disc1”. Second one is the lower disc (base) named as “disc 2” and the third
one is the curved surface of cylinder. We will now evaluate the three integrals individually.

For disc 1 and disc 2, the surface is parallel to field lines. This makes area vector perpendicular
to Electric field at all points of both the discs. Hence, both the integrals for discs become zero.

For curved surface, area vector is parallel to electric field vector. Hence angle between them is
zero at all points of curved surface. Furthermore, since all points of curved surface are
equidistant from wire, therefore, we can also assert that magnitude of electric field strength at
all points of curved surface will be same.
 ⃗⃗⃗⃗⃗ = ∫ 𝐸𝑑𝐴 cos 0 = 𝐸 ∫ 𝑑𝐴 = 𝐸2𝜋𝑟ℎ
The third integral thus becomes ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . 𝑑𝐴

∴ LHS = ∮ 𝐸 ⃗⃗⃗⃗ = ∫
⃗ . 𝑑𝐴 𝐸 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴
𝐸 ⃗⃗⃗⃗ = 0 + 0 + 𝐸2𝜋𝑟ℎ
⃗ . 𝑑𝐴
𝐸
𝐷𝑖𝑠𝑐1 𝐷𝑖𝑠𝑐2 𝑐𝑢𝑟𝑣𝑒𝑑

Step 3. Evaluation of RHS

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜆ℎ
RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

𝜆ℎ
⇒ 𝐸2𝜋𝑟ℎ =
𝜖𝑜

𝜆 2𝑘𝜆
⇒𝐸= =
2𝜋𝑟𝜖𝑜 𝑟

This is the same expression that we derived earlier using calculus.

5. Electric field strength due to a uniform distribution of charge in a cylindrical volume of

infinite length.

The case under consideration is shown in figure

A uniformly charged cylindrical region is shown. It extends up to infinity on both sides and
radius of the region is R. We wish to determine the electric field intensity due to this
arrangement of charge in its vicinity. We shall again consider the whole region in two parts.
First, region outside the cylindrical volume and second, region inside the cylindrical volume.

From symmetry, we can judge that field lines pattern will be similar , if not exactly same, to the
case of uniformly charged infinite wire.

Case 1. At a point P that lies at a distance r (r>R) from the axis of the cylindrical volume.

Step 1. We will choose a cylindrical Gaussian surface. This cylindrical surface is co-axial to given
cylindrical volume and has a radius ‘r’. Thus this surface will pass through P. Note that the
chosen surface should be a closed surface. So we will also assume two disc of radius r that will
serve as lid and base of our cylindrical surface. The height of cylindrical surface is denoted as
‘h’. So the cylindrical Gaussian surface would look as shown in figure.

Step 2. LHS = ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∫𝐷𝑖𝑠𝑐1 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝐷𝑖𝑠𝑐2 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴

In above step, we have divided the closed surface into three open surfaces. First one is upper
disc (lid) named as “Disc1”. Second one is the lower disc (base) named as “disc 2” and the third
one is the curved surface of cylinder. We will now evaluate the three integrals individually.

For disc 1 and disc 2, the surface is parallel to field lines. This makes area vector perpendicular
to Electric field at all points of both the discs. Hence, both the integrals for discs become zero.

For curved surface, area vector is parallel to electric field vector. Hence angle between them is
zero at all points of curved surface. Furthermore, since all points of curved surface are
equidistant from wire, therefore, we can also assert that magnitude of electric field strength at
all points of curved surface will be same.

⃗⃗⃗⃗⃗ = ∫ 𝐸𝑑𝐴 cos 0 = 𝐸 ∫ 𝑑𝐴 = 𝐸2𝜋𝑟ℎ

The third integral thus becomes ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . 𝑑𝐴

∴ LHS = ∮ 𝐸 ⃗⃗⃗⃗ = ∫
⃗ . 𝑑𝐴 𝐸 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴
𝐸 ⃗⃗⃗⃗ = 0 + 0 + 𝐸2𝜋𝑟ℎ
⃗ . 𝑑𝐴
𝐸
𝐷𝑖𝑠𝑐1 𝐷𝑖𝑠𝑐2 𝑐𝑢𝑟𝑣𝑒𝑑

Step 3. Evaluation of RHS

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜌𝜋𝑅 2 ℎ
RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

 𝜌𝜋𝑅 2 ℎ
⇒ 𝐸2𝜋𝑟ℎ =
𝜖𝑜

𝜌𝜋𝑅 2 2𝑘𝜆
⇒𝐸= = ,
2𝜋𝑟𝜖𝑜 𝑟

𝑤ℎ𝑒𝑟𝑒 𝜆 = 𝜌𝜋𝑅 2 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒

Case 2. At a point P that lies at a distance r (r<R) from the axis of the cylindrical volume.

Step 1. We will choose a cylindrical Gaussian surface. This cylindrical surface is co-axial to given
cylindrical volume and has a radius ‘r’. Thus this surface will pass through P. Note that the
chosen surface should be a closed surface. So we will also assume two disc of radius r that will
serve as lid and base of our cylindrical surface. The height of cylindrical surface is denoted as
‘h’. So the cylindrical Gaussian surface would look as shown in figure.
Step 2. LHS = ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 = ∫𝐷𝑖𝑠𝑐1 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝐷𝑖𝑠𝑐2 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴

In above step, we have divided the closed surface into three open surfaces. First one is upper
disc (lid) named as “Disc1”. Second one is the lower disc (base) named as “disc 2” and the third
one is the curved surface of cylinder. We will now evaluate the three integrals individually.

For disc 1 and disc 2, the surface is parallel to field lines. This makes area vector perpendicular
to Electric field at all points of both the discs. Hence, both the integrals for discs become zero.

For curved surface, area vector is parallel to electric field vector. Hence angle between them is
zero at all points of curved surface. Furthermore, since all points of curved surface are
equidistant from wire, therefore, we can also assert that magnitude of electric field strength at
all points of curved surface will be same.

⃗⃗⃗⃗⃗ = ∫ 𝐸𝑑𝐴 cos 0 = 𝐸 ∫ 𝑑𝐴 = 𝐸2𝜋𝑟ℎ

The third integral thus becomes ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . 𝑑𝐴

∴ LHS = ∮ 𝐸 ⃗⃗⃗⃗ = ∫
⃗ . 𝑑𝐴 𝐸 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴
𝐸 ⃗⃗⃗⃗ = 0 + 0 + 𝐸2𝜋𝑟ℎ
⃗ . 𝑑𝐴
𝐸
𝐷𝑖𝑠𝑐1 𝐷𝑖𝑠𝑐2 𝑐𝑢𝑟𝑣𝑒𝑑

Step 3. Evaluation of RHS

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜌𝜋𝑟 2 ℎ
RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

 𝜌𝜋𝑟 2 ℎ
⇒ 𝐸2𝜋𝑟ℎ =
𝜖𝑜
𝜌𝑟
⇒𝐸=
2𝜖𝑜
Thus the expression for E will be

𝜌𝑅 2 2𝑘𝜆
𝐸= = ,𝑟 ≥ 𝑅
2𝑟𝜖𝑜 𝑟
𝜌𝑟
𝑟 >𝑅
2𝜖𝑜

Both the cases can be represented in graph shown.

6. Electric field strength due to a uniform distribution of charge on a cylindrical surface of

infinite length.

The case under consideration is shown in figure

A uniformly charged cylindrical surface is shown. It extends up to infinity on both sides and
radius of the surface is R. We wish to determine the electric field intensity due to this
arrangement of charge in its vicinity. We shall again consider the whole region in two parts.
Firstly, the region outside the cylindrical volume and secondly, the region inside the cylindrical
surface.

From symmetry, we can judge that field lines pattern will be similar, if not exactly same, to the
case of uniformly charged infinite wire.

Case 1. At a point P that lies at a distance r (r>R) from the axis of the cylindrical surface.

Step 1. We will choose a cylindrical Gaussian surface. This cylindrical surface is co-axial to given
cylindrical surface and has a radius ‘r’. Thus this surface will pass through P. Note that the
chosen surface should be a closed surface. So we will also assume two disc of radius r that will
serve as lid and base of our cylindrical surface. The height of cylindrical surface is denoted as
‘h’. So the cylindrical Gaussian surface would look as shown in figure.

Step 2. LHS = ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∫𝐷𝑖𝑠𝑐1 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝐷𝑖𝑠𝑐2 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴
In above step, we have divided the closed surface into three open surfaces. First one is upper
disc (lid) named as “Disc1”. Second one is the lower disc (base) named as “disc 2” and the third
one is the curved surface of cylinder. We will now evaluate the three integrals individually.

For disc 1 and disc 2, the surface is parallel to field lines. This makes area vector perpendicular
to Electric field at all points of both the discs. Hence, both the integrals for discs become zero.

For curved surface, area vector is parallel to electric field vector. Hence angle between them is
zero at all points of curved surface. Furthermore, since all points of curved surface are
equidistant from wire, therefore, we can also assert that magnitude of electric field strength at
all points of curved surface will be same.

The third integral thus becomes ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∫ 𝐸𝑑𝐴 cos 0 = 𝐸 ∫ 𝑑𝐴 = 𝐸2𝜋𝑟ℎ

∴ LHS = ∮ 𝐸 ⃗⃗⃗⃗ = ∫
⃗ . 𝑑𝐴 𝐸 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴 ⃗⃗⃗⃗ + ∫
⃗ . 𝑑𝐴
𝐸 ⃗⃗⃗⃗ = 0 + 0 + 𝐸2𝜋𝑟ℎ
⃗ . 𝑑𝐴
𝐸
𝐷𝑖𝑠𝑐1 𝐷𝑖𝑠𝑐2 𝑐𝑢𝑟𝑣𝑒𝑑

Step 3. Evaluation of RHS

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜎2𝜋𝑅ℎ
RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

𝜎2𝜋𝑅ℎ
⇒ 𝐸2𝜋𝑟ℎ =
𝜖𝑜

𝜎2𝜋𝑅 2𝑘𝜆
⇒𝐸= = ,
2𝜋𝑟𝜖𝑜 𝑟

𝑤ℎ𝑒𝑟𝑒 𝜆 = 𝜎2𝜋𝑅 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

Case 2. At a point P that lies at a distance r (r<R) from the axis of the cylindrical volume.
Step 1. We will choose a cylindrical Gaussian surface. This cylindrical surface is co-axial to given
cylindrical surface and has a radius ‘r’. Thus this surface will pass through P. Note that the
chosen surface should be a closed surface. So we will also assume two disc of radius r that will
serve as lid and base of our cylindrical surface. The height of cylindrical surface is denoted as
‘h’. So the cylindrical Gaussian surface would look as shown in figure.

⃗⃗⃗⃗⃗ = ∫
Step 2. LHS = ∮ 𝐸⃗ . 𝑑𝐴 ⃗⃗⃗⃗⃗ + ∫
𝐸⃗ . 𝑑𝐴 ⃗⃗⃗⃗⃗ + ∫
𝐸⃗ . 𝑑𝐴 ⃗⃗⃗⃗⃗
𝐸⃗ . 𝑑𝐴
𝐷𝑖𝑠𝑐1 𝐷𝑖𝑠𝑐2 𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒

In above step, we have divided the closed surface into three open surfaces. First one is upper
disc (lid) named as “Disc1”. Second one is the lower disc (base) named as “disc 2” and the third
one is the curved surface of cylinder. We will now evaluate the three integrals individually.

For disc 1 and disc 2, the surface is parallel to field lines. This makes area vector perpendicular
to Electric field at all points of both the discs. Hence, both the integrals for discs become zero.

For curved surface, area vector is parallel to electric field vector. Hence angle between them is
zero at all points of curved surface. Furthermore, since all points of curved surface are
equidistant from axis, therefore, we can also assert that magnitude of electric field strength at
all points of curved surface will be same.

The third integral thus becomes ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∫ 𝐸𝑑𝐴 cos 0 = 𝐸 ∫ 𝑑𝐴 = 𝐸2𝜋𝑟ℎ

⃗ . ⃗⃗⃗⃗
∴ LHS = ∮ 𝐸 𝑑𝐴 = ∫ ⃗ . ⃗⃗⃗⃗
𝐸 𝑑𝐴 + ∫ ⃗ . ⃗⃗⃗⃗
𝐸 𝑑𝐴 + ∫ ⃗ . ⃗⃗⃗⃗
𝐸 𝑑𝐴 = 0 + 0 + 𝐸2𝜋𝑟ℎ
𝐷𝑖𝑠𝑐1 𝐷𝑖𝑠𝑐2 𝑐𝑢𝑟𝑣𝑒𝑑

Step 3. Evaluation of RHS

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 0

RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜
Step 4. LHS = RHS

0
⇒ 𝐸2𝜋𝑟ℎ =
𝜖𝑜

⇒𝐸= 0
Thus the expression for E can be written as

𝜎𝑅 2𝑘𝜆
𝐸= = ,𝑟 ≥ 𝑅
𝑟𝜖𝑜 𝑟
0 ,𝑟 < 𝑅

Both the cases can be represented in graph shown.

7. Due to a uniform distribution of charge on an infinite plane.

The situation is shown in figure. An infinite plane is shown. The surface charge density, that is ,
the charge per unit area of the sheet is denoted by σ (C/m2). The broken edges of the sheet are
meant to indicate that the sheet extends up to large distances on all sides. A point P is
considered at a perpendicular distance ‘r’ from the sheet. We intend to find the electric field
intensity at P.
We shall be using Gauss’s law to determine electric field intensity at P. As it is customary to
have an idea of field lines pattern to be able to use Gauss’s law, let us now think about it. The
field pattern must be a symmetric one as the plane extends up to infinity in all direction.
Moreover, the field must have a component away from sheet as we are considering positive
charge to be spread on the sheet. This indicates a field lines pattern as indicated in figure.
With this image of field, we start our procedure.

Step 1. We choose a cylindrical Gaussian surface. The axis of the cylinder is perpendicular to the
sheet and it passes through P. We choose the height of the cylindrical surface to be ‘2r’. So that
P lies on one of the discs that form the end of surface. See figure.
Advantages of above choice are-

1. The cylinder surface can be divided in three parts. The curved surface, the disc at left end and
the disc at right end. While the curved surface is parallel to field line, the discs are
perpendicular to field lines. Thus θ is known to us at all points of Gaussian surface. It is zero for
discs and 90o for curved surface.

Step 2. LHS = ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗

𝑑𝐴 = ∫𝐷𝑖𝑠𝑐1 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝐷𝑖𝑠𝑐2 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 + ∫𝑐𝑢𝑟𝑣𝑒𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴

In above step, we have divided the closed surface into three open surfaces. First one is left disc
named as “Disc1”. Second one is the right disc named as “disc 2” and the third one is the curved
surface of cylinder. We will now evaluate the three integrals individually.

For disc 1 and disc 2, the surface is perpendicular to field lines. Furthermore, for both discs, θ =
0o as flux is outward from both disc. Thus both these integrals will become EA cos 0.

For curved surface, surface is parallel to electric field lines. Hence θ = 90o at all points of curved
surface. Therefore, the third integral will become zero.
 ⃗ . ⃗⃗⃗⃗
∴ LHS = ∮ 𝐸 𝑑𝐴 = ∫ ⃗ . ⃗⃗⃗⃗
𝐸 𝑑𝐴 + ∫ ⃗ . ⃗⃗⃗⃗
𝐸 𝑑𝐴 + ∫ ⃗ . ⃗⃗⃗⃗
𝐸 𝑑𝐴 = 𝐸𝐴 + 𝐸𝐴 + 0
𝐷𝑖𝑠𝑐1 𝐷𝑖𝑠𝑐2 𝑐𝑢𝑟𝑣𝑒𝑑

Step 3. Evaluation of RHS

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑎𝑟𝑒𝑎 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜎𝐴

RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜

Step 4. LHS = RHS

𝜎𝐴
⇒ 2𝐸𝐴 =
𝜖𝑜

𝜎
⇒𝐸=
2𝜖𝑜

NOTE : We have done enough cases for you to notice by now that for a given type of charge
distribution, the choice of Gaussian surface remains same. Moreover, the LHS also remains
same. If you have not noticed this by now, let me elaborate.

In the first three cases ( 1. Charged particle, 2. Spherical surface, 3. Spherical volume), field
pattern was similar. It was radially outwards from a fixed point. So, in all these cases, a
spherical surface was chosen and each time we applied the Gauss’s law, the LHS would be
𝐸4𝜋𝑟 2

Similarly, in next three cases ( 4. Infinite wire, 5. Infinite cylindrical volume and 6. Infinite
cylindrical surface), the field pattern was again similar and in each of these cases, we chose a
co-axial cylindrical Gaussian surface. The LHS in each of these cases was 𝐸2𝜋𝑟ℎ

So whenever similar field pattern is given, the LHS remains unchanged. Only the RHS changes.
Keeping this observation in mind, let us see the next case.

8. Due to a uniform disribution of charge in an infinite thick layer

The case under consideration is shown in figure. There is an infinite sheet shown in figure. It is
charged uniformly throughout its volume and the volume charge density ‘ρ’ is known. The
thickness of layer is denoted by ‘a’. A plane that divides the given sheet into two sheets of
equal thickness (a/2) is termed as the mid-plane. All distances would be measured from mid-
plane. For example, if the distance of a point from mid-plane is more than a/2, then the point
lies outside the sheet and if the distance is less than a/2, then the point lies inside the layer of
charge. In this situation again, we will have to consider two cases-
1. Points outside the layer ( r > a/2 , where ‘r’ denote the distance of point from mid-plane)

2. Points inside the layer (r < a/2)

1. Electric field intensity at a point P that lies outside the layer.

The Gaussian surface would be chosen as shown in figure. The field pattern will be similar to
that of infinite plane case. Hence LHS = 2EA

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜌𝐴𝑎
RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜
 LHS = RHS

𝜌𝐴𝑎
⇒ 2𝐸𝐴 =
𝜖𝑜

𝜌𝑎
⇒𝐸=
2𝜖𝑜

2. Points inside the layer (r < a/2)

This time round, the Gaussian surface will completely lie inside the layer.

Hence LHS = 2EA

𝑞𝑒𝑛 𝑐ℎ𝑎𝑟𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑡𝑟𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝐺𝑎𝑢𝑠𝑠𝑖𝑎𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜌2𝑟𝐴
RHS = = =
𝜖𝑜 𝜖𝑜 𝜖𝑜

LHS = RHS

𝜌2𝑟𝐴
⇒ 2𝐸𝐴 =
𝜖𝑜
𝜌𝑟
⇒𝐸=
𝜖𝑜
Thus expression for E can be written as
𝜌𝑎 𝑎
𝐸= ,𝑟 ≥
2𝜖𝑜 2

𝜌𝑟
, 𝑟 < 𝑎/2
𝜖𝑜

Graph for this case will be as shown

We have now seen ample cases to familiarize you with the procedure of using Gauss’s law to
determine unknown electric field and you might have noticed some common property in all
above cases. In all the cases, charge distribution was symmetric and we were able to use this
symmetry to judge field pattern. This judgement helped us in choosing appropriate Gaussian
surface which in turn was of paramount importance to be able to evaluate LHS. It was due to
symmetry only, that we were able to pull unknown ‘E’ out of the integral sign each time. If
symmetry would have been absent, there would have been no way for us write anything
beyond “∮ 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 = ∮ 𝐸 𝑑𝐴 cos 𝜃 " .

CAUTION : Often I have seen that the students misunderstand the contents of the previous
paragraph as an assertion that “Gauss’s law is not applicable for asymmetric charge
distributions”. Be informed my dear friend, Gauss’s law is applicable to all kinds of charge
distributions. The above paragraph simply highlights our inability to evaluate LHS once
symmetry is lost.

Important examples

The ensuing examples are to be practiced and to be remembered.

Example 1 :
In a spherical region of radius R , charge is distributed non-uniformly according to the law 𝜌 =
𝐴𝑟 2 , where A is a positive constant . Find the electric field intensity at a point that lies at a
distance of ‘r’ from centre of the spherical region. Consider both cases r ≥ R and r < R.

Solution:

Since charge distribution is spherically symmetric, we should choose a concentric spherical

Gaussian surface. As explained earlier, LHS remains same for a given kind of charge distribution.
Hence LHS would be same as we obtained in case 3 of derivation given in this text. (You may
have a look at the derivation again if you are having trouble recalling the LHS.)

Therefore, LHS = 𝐸 𝑋 4𝜋𝑟 2

RHS remains to be evaluated.

Case (𝑖) 𝜌 = 𝐴𝑟 2 , where A is a positive constant

For r ≥ R, we shall find the total charge present inside spherical region of radius R as the whole
spherical region will be included in 𝑞𝑒𝑛

To get this total charge, we will have to make use of calculus as the charge density is non-
uniform.

We will consider an elemental region of width dy at a radius ‘y’ from centre of spherical region.
He charge ‘dq’ present in this elemental region will be given by 𝜌𝑑𝑉, where 𝑑𝑉 represents the
volume of this elemental region and it will be written as 𝑑𝑉 = 4𝜋𝑦 2 𝑑𝑦.
 𝑅
𝑅 𝑅 𝑦5
Thus 𝑞𝑒𝑛 = ∫ 𝑑𝑞 = ∫ 𝜌𝑑𝑉 = ∫0 𝜌4𝜋𝑦 2 𝑑𝑦 = ∫0 𝐴𝑦 2 4𝜋𝑦 2 𝑑𝑦 = 4𝜋𝐴 [ 5 ] = 4𝜋𝐴𝑅 5 /5
0

Thus RHS becomes 4𝜋𝐴𝑅 5 /5𝜖𝑜

𝐿𝐻𝑆 = 𝑅𝐻𝑆
2
4𝜋𝐴𝑅 5
∴ 𝐸 𝑋 4𝜋𝑟 =
5𝜖𝑜

𝐴𝑅5
⇒𝐸= ,𝑟 ≥ 𝑅
5𝜖𝑜 𝑟 2

For r < R , the LHS will again remain same. Only RHS will change. This time while evaluating 𝑞𝑒𝑛 ,
the entire charge will not included as Gaussian surface is smaller as compared to region in
which charge is present. So charge only up to radius ‘r’ will be included. Thus we should simply
change the limits of the integral from 0 to ‘r’

𝑟
𝑟 𝑟 𝑦5
Thus 𝑞𝑒𝑛 = ∫ 𝑑𝑞 = ∫ 𝜌𝑑𝑉 = ∫0 𝜌4𝜋𝑦 2 𝑑𝑦 = ∫0 𝐴𝑦 2 4𝜋𝑦 2 𝑑𝑦 = 4𝜋𝐴 [ 5 ] = 4𝜋𝐴𝑟 5 /5
0

Thus RHS becomes 4𝜋𝐴𝑟 5 /5𝜖𝑜

𝐿𝐻𝑆 = 𝑅𝐻𝑆
 2
4𝜋𝐴𝑟 5
∴ 𝐸 𝑋 4𝜋𝑟 =
5𝜖𝑜

𝐴𝑟3
⇒𝐸= ,𝑟 < 𝑅
5𝜖𝑜

AN ADVICE: You will come across many such problems in which a varying charge density will be
given to you in a spherical region and you will be asked to determine electric field intensity in
both the regions, the inner and the outer region. Remember, the entire procedure will remain
as it is, just the expression used in the two integrals ( while evaluating 𝑞𝑒𝑛 ) will change. Same is
true for cylindrical charge distribution also. See the next example.

Example :

In a cylindrical region of radius R and of infinite length, charge is non-uniformly distributed

𝑟
according to the law 𝜌 = 𝜌𝑜 (𝑎 − 𝑏) , where 𝜌𝑜 , 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠.Find electric
field intensity due to this charge distribution.

Solution:

For a cylindrical charge distribution, LHS will be 𝐸 𝑥 2𝜋𝑟ℎ.

For RHS, we will consider an elemental cylindrical region of radius ‘y’ and thickness ’dy’ and of
height ‘h’.
𝑦
𝑞𝑒𝑛 = ∫ 𝑑𝑞 = ∫ 𝜌𝑑𝑉 = ∫ 𝜌2𝜋𝑦ℎ𝑑𝑦 = ∫ 𝜌𝑜 (𝑎 − 𝑏 ) 2𝜋𝑦ℎ𝑑𝑦

For 𝑟 ≥ 𝑅 , limits will be from 0 to R .

𝑅
𝑅 𝑦 𝑎𝑦 2 𝑦3 𝑎𝑅 2 𝑅3
∴ 𝑞𝑒𝑛 = ∫0 𝜌𝑜 (𝑎 − 𝑏 ) 2𝜋𝑦ℎ𝑑𝑦 = 2𝜋ℎ𝜌𝑜 [ − 3𝑏] = 2𝜋ℎ𝜌𝑜 [ − 3𝑏 ]
2 0 2

𝑎𝑅 2 𝑅3
⇒ 𝐸 𝑥 2𝜋𝑟ℎ = 2𝜋ℎ𝜌𝑜 [ − 3𝑏 ] /𝜖𝑜
2

𝑎𝑅 2 𝑅3
𝜌𝑜 [ − 3𝑏 ]
2
⇒𝐸= ,𝑟 ≥ 𝑅
𝑟𝜖𝑜

For 𝑟 < 𝑅 , limits will be from 0 to r .

 𝑟
𝑟 𝑦 𝑎𝑦 2 𝑦3 𝑎𝑟 2 𝑟3
∴ 𝑞𝑒𝑛 = ∫0 𝜌𝑜 (𝑎 − 𝑏 ) 2𝜋𝑦ℎ𝑑𝑦 = 2𝜋ℎ𝜌𝑜 [ − 3𝑏] = 2𝜋ℎ𝜌𝑜 [ − 3𝑏]
2 0 2

𝑎𝑟 2 𝑟3
⇒ 𝐸 𝑥 2𝜋𝑟ℎ = 2𝜋ℎ𝜌𝑜 [ − 3𝑏] /𝜖𝑜
2

𝑎𝑟 𝑟2
𝜌𝑜 [ − ]
2 3𝑏
⇒𝐸= ,𝑟 < 𝑅
𝜖𝑜

A special case : Uniformly charged spherical region with a spherical cavity

The figure shows a spherical region having a uniform density with a spherical cavity in it. ‘C’ is
the centre of the spherical region and ‘O’ is the centre of the cavity. We wish to find the electric
field intensity at any general point P inside the cavity.

Our approach would be to first write the electric field intensity vector at P assuming that the
spherical region is completely filled with charge and then we will subtract the contribution in
electric field intensity due to the charge that has been removed to make cavity.

𝐸⃗𝑃 = 𝐸⃗𝑑𝑢𝑒 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑠𝑝ℎ𝑒𝑟𝑒 − 𝐸⃗𝑑𝑢𝑒 𝑡𝑜 𝑟𝑒𝑚𝑜𝑣𝑒𝑑 𝑝𝑎𝑟𝑡

⃗⃗⃗⃗⃗
𝜌𝐶𝑃 ⃗⃗⃗⃗⃗
𝜌𝑂𝑃 ⃗⃗⃗⃗⃗
𝜌𝐶𝑂
⇒ 𝐸⃗𝑃 = − =
3𝜖𝑜 3𝜖𝑜 3𝜖𝑜
The significant aspect about this result is that 𝐸⃗𝑃 is independent of P! This suggests that the
cavity will be a region of uniform electric field. This uniform field will be directed along line
𝜌 𝑥 𝐶𝑂
joining C and O. And its magnitude will be
3𝜖𝑜

The figure shown indicates the field lines in the cavity.