Contents
Physics Topics can be both theoretical and experimental, with scientists using a range of tools and techniques to understand the phenomena they investigate.
Statement and Proof of Gauss’ Theorem
Statement:
The net electric flux linked with a closed surface is \(\frac{1}{\epsilon}\) times the net charge within the surface.
Mathematically, ϕ = \(\oint_S \vec{E} \cdot d \vec{S}\) = \(\frac{q}{\epsilon}\) …… (1)
Here q is the total charge enclosed by the surface and ε is the permittivity of the medium.
For vacuum, ϕ = \(\oint_S \vec{E} \cdot d \vec{S}\) = \(\frac{q}{\epsilon_0}\)
Proof:
Let S be a closed surface enclosing +q charge situated at O [Fig.], P is a point situated on the surface and dS is a small area on the surface surrounding P. Suppose, OP = r.
Electric field intensity at P due to the charge +q,
E = \(\frac{1}{4 \pi \epsilon} \cdot \frac{q}{r^2}\)
Let \(\overrightarrow{P N}\) be the normal at P on dS and θ be the angle between \(\vec{E}\) and \(\overrightarrow{P N}\).
∴ Electric Flux through area dS
= \(\vec{E} \cdot d \vec{S}\) = EdS cos θ
= \(\frac{1}{4 \pi \epsilon} \cdot \frac{q}{r^2} d S \cos \theta\) = \(\frac{q}{4 \pi \epsilon} \cdot \frac{d S \cos \theta}{r^2}\) = \(\frac{q}{4 \pi \epsilon} \cdot d \omega\)
Here, dω = \(\frac{d S \cos \theta}{r^2}\) = solid angle subtended at O by the area dS.
Therefore, net flux linked with the entire closed surface,
[because a closed surface subtends a solid angle 4π at any point within the surface]
If q1, q2, ….., qn be the charges lying within the closed surface and E1, E2,…….., En be the corresponding electric fields, the net flux linked with the closed surface will be,
where Q is the net charge inside the closed surface.
According to the nature of the charges (positive and negative) positive or negative signs are to be used in the sum of the charges.
Electric flux due to a charge lying outside a closed surface: Let a charge of +q be placed at O outside the closed surface S [Fig.]. A cone of solid angle dω having its vertex at O intercepts small areas dS1 and dS2 on the closed surface. Let P be a point on dS1 and Q be another point on dS2. E1 and E2 are the electric field intensities at P and Q, respectively.
Suppose OP = r1 and OQ = r2.
Component of E1 along the normal to dS1 = E1cosθ1 and component of E2 along the normal to dS2 = E2 cosθ2. Since electric flux through dS1 directed inwards is negative, the flux through dS2 directed outwards is positive.
Therefore, electric flux through dS1 = -E1 cosθ1dS1 and electric flux through dS2 = E2cosθ2dS2
So, net electric flux through dS1 and dS2
Clearly, this result is in accordance with equation (1), because in this case, q = charge within the closed surface = 0.
Discussions:
- Only the charges inside a volume contribute to the electric flux linked with the surface enclosing that volume.
- The net charge q is the algebric sum of all the charges, pos-itive or negative, inside the closed surface.
- If the net charge q is positive, the outward flux across the closed surface is greater than the inward flux; the enclosed volume acts as a source.
- If the net charge q is negative, the inward flux exceeds the outward flux; the enclosed volume acts as a sink.
- If the dosed surface encloses equal amounts of positive and negative charges, then the net charge q is zero; conse-quently, the net flux linked with the surface is also zero.
For example, if a closed surface encloses an electric dipole, the net electric flux through it is zero. - Charges outside the closed surface have no net contribution towards the electric flux linked with it.
Applications of Gauss’ Theorem
Field intensity at a point due to a point charge: we have to calculate the electric field intensity at a point situated at a distance r from a point charge q . Let us consider a spherical Gaussian surface of radius r having the point charge q at its centre [Fig.]. We can say from symmetry that electric field intensity due to a positive point charge is directed radially out-wards. So intensity at any point on the Gaussian surface is perpendicular to it. Therefore, \(\vec{E}\) and d\(\vec{S}\) are parallel at each point.
where k and ε0 are respectively the dielectric constant of the medium and the permittivity of vacuum.
In vector form, \(\vec{E}\) = \(\frac{q}{4 \pi \kappa \epsilon_0 r^2} \hat{r}\)
where \(\hat{r}\) is a unit vector along the outward drawn normal.
Note that, if the point charge is negative, q is replaced by -q. Then E would be negative, meaning that it would be radially inwards.
Field intensity at a point due to a uniformly charged thin spherical shell:
Point outside the spherical shell: Suppose, a thin spherical shell of radius R is charged uniformly with +q [Fig.],
Electric field intensity at any external point P is to be calculated. Suppose, the distance of P from the centre O of the shell is r. Imagine a Gaussian spherical shell of radius r with centre at O.
We can say from symmetry that electric field intensity at each point on the surface of the spherical shell is equal and is directed along the outward drawn normal.
Suppose, intensity at P is E. Electric flux linked with the imagi-nary surface of the spherical shell = E ᐧ 4πr2.
Since the Gaussian surface encloses the charge +q, the net flux according to Gauss’ theorem is \(\frac{q}{\epsilon}\).
∴ E ᐧ 4πr2 = \(\frac{q}{\epsilon}\)
or, E = \(\frac{1}{4 \pi \epsilon} \cdot \frac{q}{r^2}\) = – e 1 ± 4ne r2
In vector form, \(\vec{E}\) = \(\frac{1}{4 \pi \epsilon} \cdot \frac{q}{r^2} \hat{r}\)
where \(\hat{r}\) is a unit vector along the outward drawn normal. Equation (2) shows that for an external point, a charged spher-ical shell behaves as though its total charge were concentrated at its centre.
Point inside the spherical shell: The point p is situated inside the charged spheical shell [Fig.], Imagine a Gaussian spherical shell of radius r with centre at O .
Since the Gaussian surface does not enclose any charge (because the charge lies on the surface of the spherical shell), the flux linked with the imaginary surface of the spherical shell is given by,
E ᐧ 4πr2 = 0 [by Gauss’ theorem]
or, E = 0
i. e., there exists no field intensity inside a charged spherical shell.
Field intensity at a point due to an infinitely long straight charged conducting wire: Consider an infinitely long thin straight wire charged uniformly [Fig.],
Let +λ be the linear charge density, i.e., charge per unit length of the wire. Electric field intensity is to be calculated at P at a distance r from the wire. Imagine a cylinder of radius r and length l with the wire as its axis. Point P lies on the surface of such a cylinder. The surface of this cylinder will act as the Gaussian surface. All the points on the curved surface of this cylinder are equidistant from the wire.
From symmetry, it can be said that electric field intensity at each point on the curved surface of the imaginary cylinder is equal and normally outwards. No electric flux is linked with the two flat circular faces of the cylinder as the direction of electric field E is parallel to these two faces (i.e., the component of the electric field along the normal to the two flat circular faces is zero).
So, electric flux linked with the curved surface of the Gaussian cylinder = E ᐧ 2πrl.
Now, since the cylinder encloses an amount of charge λl, by Gauss’ theorem, electric flux linked with the Gaussian cylinder = \(\frac{\lambda l}{\epsilon}\)
∴ E ᐧ 2πrl = \(\frac{\lambda l}{\epsilon}\) or, E = \(\frac{1}{4 \pi \epsilon} \cdot \frac{2 \lambda}{r}\) ….. (3)
In vector form, E = \(\frac{1}{4 \pi \epsilon} \cdot \frac{2 \lambda}{r} \hat{r}\)
where \(\hat{\gamma}\) is a unit vector along the outward drawn normal.
Field intensity at a point due to an infinite noncon-ducting uniformly charged plane lamina: Consider an infinite nonconducting uniformly charged plane lamina having a surface charge density +σ [Fig.].
A point P is taken in front of the plane lamina, where the electric field intensity is to be determined. A small cylinder of cross sectional area dS, normal to the lamina and having equal lengths on either side of it, is imagined such that P and P’ lie on its flat circular faces. The surface of this cylinder will act as the Gaussian surface. Since the lamina is infinite, the electric field intensity at each point of the two flat circular faces of the Gaussian cylinder is equal and normally outwards due to symmetry.
Electric flux through the curved surface of the Gaussian cylinder is zero, since the direction of E is parallel to the curved surface (i.e., the component of E along the normal to the curved surface at every point is zero).
Now, electric flux through the two flat circular faces of the Gaussian cylinder = E ᐧ 2 dS
Now, since the cylinder encloses an amount of charge σdS, by Gauss’ theorem, electric flux linked with the cylinder = \(\frac{\sigma d S}{\epsilon}\)
∴ E ᐧ 2 dS = \(\frac{\sigma d S}{\epsilon}\) or E = \(\frac{\sigma}{2 \epsilon}\) = \(\frac{\sigma}{2 \kappa \epsilon_0}\) ….. (4)
In vector form, \(\vec{E}\) = \(\frac{\sigma}{2 \epsilon} \hat{n}\) = \(\frac{\sigma}{2 \kappa \epsilon_0} \hat{n}\)
where \(\hat{n}\) is the unit vector directed normally away from the plane.
Field intensity at a point due to a charged conductor: Let C be a charged conductor [Fig]. The electric field intensity, developed due to it, at an external point P is to be determined. The distance of P from the conductor is much less than the dimensions of the conducting body.
Now, we consider a right circular cylinder of cross sectional area dS, normal to the surface of the conductor, one of whose flat circular faces contains the point P. The surface of this cylinder serves as the Gaussian surface in this case. We assume that the cross sectional area dS is small enough that E is constant over the flat circular face. From symmetry, the electric field intensity is normally outwards with respect to the surface dS.
Thus, electric flux across the flat circular face around P is Eds.
The curved surface of the cylinder does not contribute to any electric flux, since the direction of E is parallel to the curved surface at every point (i.e., the component of E along the nor-mal to the curved surface at every point is zero). Also, the flat circular face inside the conducting body has no contribution, because the electric field inside a conductor is zero.
The entire charge resides on the outer surface of the conductor. In general, the curvature is not uniform around the surface. As a result, the charge would be non-uniformly distributed.
Let +σ be the surface charge density on a small area dS on the surface just in front of P.
Now, since the cylinder encloses an amount of charge crdS, by Gauss’ theorem, electric flux linked with the Gaussian cylinder = \(\frac{\sigma d S}{\epsilon}\)
∴ EdS = \(\frac{\sigma d S}{\epsilon}\) or, E = \(\frac{\sigma}{\epsilon}\) ……….. (5)
In vector form, \(\vec{E}\) = \(\frac{\sigma}{\epsilon_0} \hat{n}\)
where \(\hat{n}\) is the unit vector normal to the surface in the outward direction.
Numerical Examples
Example 1.
An electric field (2\(\hat{\boldsymbol{i}}\) + 3\(\hat{\boldsymbol{j}}\)) N ᐧ C-1 exists in a region. Cal-culate the electric flux linked with a square plate of side 0.5 m held parallel to
(i) yz -plane and
(ii) xy -plane.
Solution:
Electric field, \(\vec{E}\) = 2\(\hat{\boldsymbol{i}}\) + 3\(\hat{\boldsymbol{j}}\) ; so it lies on the xy -plane.
Area of the square plate, S = (0.5)2 = 0.25 m2
i) The vector representation of the plate parallel to the yz -plane is, \(\vec{S}\) = S\(\hat{\boldsymbol{i}}\) = 0.25\(\hat{\boldsymbol{i}}\)
∴ Electric flux linked with the square plate
= \(\vec{E}\) ᐧ \(\vec{S}\) = (2\(\hat{\boldsymbol{i}}\) + 3\(\hat{\boldsymbol{j}}\)) ᐧ (0.25\(\hat{\boldsymbol{i}}\)) = 0.5 N ᐧ m2 ᐧ C,sup>-1
ii) The vector representation of the plate parallel to the xy -plane is, \(\vec{S}\) = S\(\hat{\boldsymbol{k}}\) = 0.25\(\hat{\boldsymbol{k}}\)
∴ Electric flux linked with the square plate
= \(\vec{E}\) ᐧ \(\vec{S}\) = (2\(\hat{\boldsymbol{i}}\) + 3\(\hat{\boldsymbol{j}}\)) ᐧ (0.25\(\hat{\boldsymbol{k}}\)) = 0.
Example 2.
In vacuum,
i) find out the electric flux across an area \(\overrightarrow{\boldsymbol{S}}\) = 10\(\hat{j}\) placed in an electric flux \(\overrightarrow{\boldsymbol{E}}\) = 2\(\hat{\boldsymbol{i}}\) + 4\(\hat{\boldsymbol{j}}\) + 7\(\hat{\boldsymbol{k}}\).
ii) How many electric charges are to be placed at a point such that 4400 number of electric lines of force will emerge from that point?
iii) What will be the electric flux through any one of the faces of a cube of side 10 cm if a charge of 1 eus is placed at the centre of the cube?
Solution:
i) Electric flux, ϕ= \(\vec{E} \cdot \vec{S}\) = (2\(\hat{\boldsymbol{i}}\) + 4\(\hat{\boldsymbol{j}}\) + 7\(\hat{\boldsymbol{k}}\)) ᐧ 10j = 40 unit
ii) In CGS system, according to Gauss’ theorem, in vacuum the net electric flux linked with a closed surface is 4π times the net charge within the surface.
∴ 4400 = 4πq
or, q = \(\frac{4400}{4 \pi}\) = \(\frac{4400 \times 7}{4 \times 22}\) = 350 esu
iii) Net electric flux =4πq = 4π ᐧ 1 = 4π esu
As a cube has 6 faces, the electric flux linked with each face of the cube = \(\frac{1}{6}\) × 4π = \(\frac{2}{3}\)π esu.
Example 3.
An electric flux of 6.5 × 103 N m2 C-1 is linked with a sphere due to some charge placed in vacuum inside the sphere. Calculate the magnitude of the charge.
Solution:
In this case, total flux linked = 6.5 × 103 N ᐧ m2 ᐧ C-1
According to Gauss’ theorem,
total flux linked with the sphere = \(\frac{1}{\epsilon_0} \cdot q\)
∴ \(\frac{1}{\epsilon_0} \cdot q\) = 6.5 × 103
or, q = 6.5 × 103 × ε0
= 6.5 × 103 × 8.85 × 10-12
= 5.75 × 10-8C
Example 4.
A spherical shell of radius 20cm has 20μC charge placed in vacuum. Calculate the electric field intensity
(i) at a distance of 15cm and
(ii) at a distance of 40 cm from the centre of the spherical shell.
Solution:
i) Electric field intensity at a distance of 15 cm from the centre of the spherical shell = 0; because charge does not reside inside a charged spherical shell.
ii) Electric field intensity at a distance of 40 cm or 0.4 m from the centre of the spherical shell,
E = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{x^2}\) = \(\frac{9 \times 10^9 \times 20 \times 10^{-6}}{(0.4)^2}\)
= 1.125 × 106 N ᐧ C-1
Example 5.
A thin straight wire of length 40 cm placed in vacuum has 20µC charge. Calculate the electric field intensity at a distance 15 cm from the wire.
Solution:
Electric field intensity at a distance of 15cm from the straight wire,
E = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \lambda}{x}\) [λ = charge per unit length of the wire = \(\frac{20 \times 10^{-6}}{40 \times 10^{-2}}\)C ᐧ m-1]
= \(\frac{9 \times 10^9 \times 2 \times 20 \times 10^{-6}}{15 \times 10^{-2} \times 40 \times 10^{-2}}\) = 6 × 106 N ᐧ C-1
Example 6.
A cube placed in vacuum contains a charge of 9 × 10-9 C. What will be the electric flux linked with each face of the cube?
Solution:
Electric flux linked with the 6 faces of the cube = \(\frac{q}{\epsilon_0}\)
Due to symmetry, this flux is equally shared among the 6 faces.
So, the electric flux linked with each face of the cube
= \(\frac{1}{6} \cdot \frac{q}{\epsilon_0}\) = \(\frac{1}{6} \times \frac{9 \times 10^{-9}}{8.854 \times 10^{-12}}\)
= 169.41 V ᐧ m
Example 7.
An electric field is expressed as \(\overrightarrow{\boldsymbol{E}}\) = (5\(\hat{\boldsymbol{i}}\) + 3\(\hat{\boldsymbol{j}}\) + 2\(\hat{\boldsymbol{k}}\)) unit. Find out the electric flux across an area of 200 unit on the yz -plane in that field. [HS ‘13]
Solution:
The yz -plane is perpendicular to the x -axis.
∴ The given area vector is \(\vec{S}\) = 200\(\hat{i}\) unit.
Then, the electric flux across that area,
ϕ = \(\vec{E} \cdot \vec{S}\) = (5\(\hat{i}\) + 3\(\hat{j}\) + 2\(\hat{k}\)) = 1000 unit.
Example 8.
The surface density of charge on a large vertical positively charged plate is σ C ᐧ m-2. A string attaches a metal ball of mass M and charge +q with the plate.
Find out the angle between the string and the plate in equilibrium. [Hs ‘13]
Solution:
Let the angle between the string and the plate in equilibrium = θ [Fig.].
Tension in the string = T;
Electric field at the position of the ball = E
= \(\frac{\sigma}{2 \epsilon_0}\)
In equilibrium,
T sin θ = qE = \(\frac{q \sigma}{2 \epsilon_0}\); T cos θ = Mg
∴ tan θ = \(\frac{q \sigma}{2 \epsilon_0 M g}\)
or, θ = tan–\(\left(\frac{q \sigma}{2 \epsilon_0 M g}\right)\)
Example 9.
The classical concept of atomic structure is that negative charges are uniformly distributed inside a sphere of radius R keeping the nucleus of positive charge Ze at the centre of that sphere. An atom as a whole is electrically neutral. Find out the electric field at a distance
r from the nucleus, according to this atomic model. [NCERT]
Solution:
Charge of the nucleus = +Ze.
∴ Amount of charge outside the nucleus = —Ze, as an atom is electrically neutral.
Volume density of negative charge inside the sphere,
ρ = \(\frac{-Z e}{\frac{4}{3} \pi R^3}\) = \(-\frac{3 Z e}{4 \pi R^3}\)
Electric field at external points (r> R): Let A be any such point outside the sphere [Fig.]. A concentric sphere of radius r, passing through A, is a convenient Gaussian surface.
The charge enclosed by this Gaussian surface,
q = charge of the nucleus + charge distributed inside the sphere of radius R
= +Ze + (-Ze) = 0
∴ From Gauss’ theorem, \(\oint_S \vec{E} \cdot d \vec{S}\) = \(\frac{q}{\epsilon_0}\) = 0
∴ E = 0, at all external points.
Electric field at internal points (r < R) : Let B be any such point inside the sphere. A concentric sphere of radius r, passing through B, is a convenient Gaussian surface [Fig.], It is to be noted that this surface encloses the entire positive charge +Ze of the nucleus, but a part of the negative charge -Ze remains outside it.
The amount of negative charge inside this Gaussian surface,
q’ = volume of this enclosed sphere × volume density ρ of negative charge
= \(\frac{4}{3} \pi r^3\left(-\frac{3 Z e}{4 \pi R^3}\right)\) = -Ze\(\frac{r^3}{R^3}\)
∴ Net charge enclosed by the sphere,
q = +Ze + q’ = Ze – Ze\(\frac{r^3}{R^3}\) = Ze(1 – \(\frac{r^3}{R^3}\))
For any surface element \(d \vec{S}\) on the Gaussian surface, the electric field \(\vec{E}\) and the area vector \(d \vec{S}\) are parallel to each other.
∴ \(\vec{E} \cdot d \vec{S}\) = EdS cos 0° = EdS
Again, from symmetry, E is uniform in magnitude over the Gaussian surface. Then, from Gauss’ theorem,
Example 10.
A uniform electric field along the x – axis is given as,
\(\vec{E}\) = (200\(\hat{i}\)) N ᐧ C-1, for x > 0
= (-200\(\hat{i}\)) N ᐧ C-1, for x < 0
A cylinder of length 20 cm and radius 5 cm has its centre at the origin and axis along the x -axis is placed in vacuum. Find out
(i) the electric flux across each of its circular faces,
(ii) the flux across its curved surface,
(iii) the flux across its entire outer surface and
(iv) the net charge enclosed by it. [NCERT]
Solution:
Radius of the cylinder, r = 5 cm = 0.05 m
∴ Area of each circular face,
S = πr2 = 3.14 × (0.05)2 m2
As the length of the cylinder is 20 cm or 0.2 m, the two circular faces are at x = +0.1 m and x = -0.1 m.
The area vectors representing the right and the left circular faces [Fig.] are, \(\vec{S}_1\) = \(\hat{i}\left(\pi r^2\right)\) and \(\vec{S}_2\) = \(-\hat{i}\left(\pi r^2\right)\) respectively.
The electric fields at the positions of these circular faces are, \(\vec{E}_1\) = \((200 \hat{i})\) N ᐧ C-1 and \(\vec{E}_2\) = \((-200 \hat{i})\) N ᐧ C-1 respectively.
i) The electric flux across the right circular face,
ϕ1 = \(\vec{E}_1 \cdot \vec{S}_1\) = \((200 \hat{i}) \cdot \hat{i}\left(\pi r^2\right)\)
= 200 × 3.14 × (0.05)2 = 1.57 N ᐧ m2 ᐧ C-1
Similarly, for the left circular face,
ϕ2 = \(\vec{E}_2 \cdot \vec{S}_2\) = \((-200 \hat{i}) \cdot\left(-\hat{i} \pi r^2\right)\)
= 200 × 3.14 × (0.05)2 = 1.57 N ᐧ m2 ᐧ C-1
∴ ϕ1 = ϕ2 i.e., equal flux passes across each of the two faces.
ii) The curved surface is everywhere parallel to the electric field vector.
∴ The electric flux linked with the curved surface = 0
iii) The entire outer surface consists of the two circular faces and the curved surface. So, the flux linked with the entire surface,
∴ ϕ = ϕ1 + ϕ2 + 0 = 1.57 + 1.57 + 0
= 3.14 N ᐧ m2 ᐧ C-1
iv) From Gauss’ theorem, net flux, ϕ = \(\frac{q}{\epsilon_0}\). So the enclosed charge is,
q = ϕε0 = 3.14 × 8.854 × 10-12
= 2.78 × 10-11 C