Physics ELECTRIC DIPOLE AND GAUSS LAW

### TOPIC'S COVERED

color{blue}{star} Electric Dipole
color{blue}{star} The field of an Electric Dipole
color{blue}{star} Dipole in a Uniform External Field
color{blue}{star} Continuous Charge Distribution
color{blue}{star} Gauss's Law
color{blue}{star} Application of Gauss's law

### Electric Dipole

\color{blue}✍️ An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a.

\color{blue}✍️ The line connecting the two charges defines a direction in space.

\color{blue}✍️ The direction from –q to q is said to be the direction of the dipole.

\color{blue}✍️ The mid-point of locations of –q and q is called the centre of the dipole.

\color{blue}✍️ The total charge of the electric dipole is obviously zero. This does not mean that the field of the electric dipole is zero. Since the charge q and –q are separated by some distance, the electric fields due to them, when added, do not exactly cancel out. However, at distances much larger than the separation of the two charges forming a dipole (r > > 2a), the fields due toq and  –q nearly cancel out. The electric field due to a dipole therefore falls off, at large distance, faster than like 1/r^2 .

### The field of an electric dipole

The electric field of the pair of charges (–q and q) at any point in space can be found out from Coulomb’s law and the superposition principle.

\color{blue} ✍️ The results are simple for the following two cases:

color{blue} {(i)} When the point is on the dipole axis,

color{blue} {(ii)} When it is in the equatorial plane of the dipole, i.e., on a plane perpendicular to the dipole axis through its centre.

\color{blue} ✍️The electric field at any general point P is obtained by adding the electric fields E_(–q) due to the charge –q and E_(+q) due to the charge q, by the parallelogram law of vectors.

color{blue}{(i)} For points on the axis
Let the point P be at r from the centre of the dipole on the side of the charge q. Then

E_(-q) =- q/(4πε_0(r+a)^2)hat p

where hat p is the unit vector along the dipole axis (from –q to q) . Also

E_(+q) = q/(4πε_0(r-a)^2)hat p

The total field at P is

E =E_(+q)+E_(-q)= q/(4πε_0)[1/(r-a)^2-1/(r+a)^2]hat p

= q/(4πε_0)(4ar)/(r^2-a^2)^2

\color{blue} ✍️For r > > a

E=(4qa)/(4πε_0r^3) hat p

color{blue}{(ii)} For points on the equatorial plane :

The magnitudes of the electric fields due to the two charges +q and –q are given by

E_(+q) = q/(4πε_0(r-a)^2)hat p and

E_(-q) =- q/(4πε_0(r+a)^2)hat p

are equal.

color{blue}●The components normal to the dipole axis cancel away.

color{blue}●The components along the dipole axis add up. The total electric field is opposite to hat p. We have

E = - (E_(+q) + E_(–q) ) cosθ hat p

=- (2qa)/(4πε_0(r^2+a^2)^(3/2)) hat p

At large distances (r > > a), this reduces

E=- (2qa)/(4πε_0r^3) hatp

color{blue}● The dipole field at large distances does not involve q and a separately;

color{blue}● It depends on the product qa.

The dipole moment vector p of an electric dipole is defined by p = q × 2a hat p

color{blue}● It is a vector whose magnitude is charge q times the separation 2a.

color{blue}● The direction is along the line from –q to q. In terms of p.

color{blue}● The electric field of a dipole at large distances takes simple forms :

color{blue}{1.} At a point on the dipole axis:

E=(2P)/(4πε_0r^3) (r >> a)

color{blue}{2.} At a point on the equatorial plane

E=(-P)/(4πε_0r^3) (r >> a)

color{blue}● The dipole field at large distances falls off not as 1/r^2 but as 1/r^3.

color{blue}● The magnitude and the direction of the dipole field depends not only on the distance r but also on the angle between the position vector r and the dipole moment p.

color{blue}● When the dipole size 2 a approaches zero, the charge q approaches infinity in such a way that the product p = q × 2 a is finite. Such a dipole is referred to as a point dipole.

### Physical Significance of Dipoles

color{blue}{1.} In most molecules, the centres of positive charges and of negative charges lie at the same place. Therefore, their dipole moment is zero. CO_2  and CH_4 are of this type of molecules. However, they develop a dipole moment when an electric field is applied.

color{blue}{2.} In some molecules, the centres of negative charges and of positive charges do not coincide. Therefore they have a permanent electric dipole moment, even in the absence of an electric field. Such molecules are called polar molecules. Water molecules, H_2O, is an example of this type.
Q 3270680516

Two charges ±10 μC are placed 5.0 mm apart.
Determine the electric field at (a) a point P on the axis of the dipole
15 cm away from its centre O on the side of the positive charge, as
shown in Fig. 1.21(a), and (b) a point Q, 15 cm away from O on a line
passing through O and normal to the axis of the dipole, as shown in
Fig. 1.21(b).
Class Chapter 1 Example 10
Solution:

(a) Field at P due to charge +10 μC

= ( 10^(-5) c)/( 4 pi (8.854 xx 10^(-12) C^2 N^(-1) m^(-2))) xx 1/((15 - 0.25)^2 xx 10^(-4) m^2)

= 4.13 xx 10^6 N C^(–1) along BP

Field at P due to charge –10 μC

 = ( 10^(-5) c)/( 4 pi (8.854 xx 10^(-12) C^2 N^(-1) m^(-2))) xx 1/((15 + 0.25)^2 xx 10^(-4) m^2)

= 3.86 xx 10^6 N C^(–1) along PA

The resultant electric field at P due to the two charges at A and B is

= 2.7 xx 10^5 N C^(–1)  along BP.
In this example, the ratio OP/OB is quite large (= 60). Thus, we can
expect to get approximately the same result as above by directly using
the formula for electric field at a far-away point on the axis of a dipole.
For a dipole consisting of charges ± q, 2a distance apart, the electric
field at a distance r from the centre on the axis of the dipole has a
magnitude

 E = (2p)/( 4 pi epsilon_0 r^3)

where p = 2a q is the magnitude of the dipole moment.
The direction of electric field on the dipole axis is always along the
direction of the dipole moment vector (i.e., from –q to q). Here,
p =10^(–5) C xx 5 xx 10^(–3) m = 5 xx 10^(–8) C m

Therefore,

 E = (2 xx 5 xx 10^(-8) Cm)/( 4 pi (8.854 xx 10^(-12) C^2 N^(-1) m^(-2))) xx 1/( (15)^3 xx 10^(-6) m^3) = 2.6 xx 10^5 N C^(–1)

along the dipole moment direction AB, which is close to the result
obtained earlier.
(b) Field at Q due to charge + 10 μC at B

 = ( 10^(–5) C) /( 4 pi (8.854 xx 10^(-12) C^2 N^(-1) m^(-2))) xx 1/( [15^2 + (0.25)^2 ] xx 10^(-4) m^2)

= 3.99 xx 10^6 N C^(–1) along BQ

Field at Q due to charge –10 μC at A

= ( 10^(-5) C)/( 4 pi (8.854 xx 10^(-12) C^2 N^(-1) m^(-2))) xx 1/( [15^2 + (0.25)^2 ] xx 10^(-4) m^2)

= 3.99 xx 10^6 N C^(–1) along QA.

Clearly, the components of these two forces with equal magnitudes
cancel along the direction OQ but add up along the direction parallel
to BA. Therefore, the resultant electric field at Q due to the two
charges at A and B is

 = 2 xx (0.25)/sqrt( 15^2 + (0.25)^2) xx 3.99 xx 10^6 N C^(-1) along BA

= 1.33 xx 10^5 N C^(–1) along BA.

As in (a), we can expect to get approximately the same result by
directly using the formula for dipole field at a point on the normal to
the axis of the dipole:

 E = p/(4 pi epsilon_0 r^3)

 = ( 5 xx 10^(-8) Cm)/( 4 pi (8.854 xx 10^(-12) C^2 N^(-1) m^(-2))) xx 1/( (15)^3 xx 10^(-6) m^3)

 = 1.33 xx 105 N C^(–1).

The direction of electric field in this case is opposite to the direction
of the dipole moment vector. Again the result agrees with that obtained
before.

### DIPOLE IN A UNIFORM EXTERNAL FIELD

Consider a permanent dipole of dipole moment p in a uniform external field E. There is a force qE on q and a force –qE on –q. The net force on the dipole is zero, since E is uniform. When the net force is zero, the torque is independent of the origin. Its magnitude equals the magnitude of each force multiplied by the arm of the couple.

Magnitude of torque = q E × 2 a sinθ

= 2 q a E sinθ

Its direction is normal to the plane of the paper, coming out of it.

The magnitude of p × E  is also p E sinθ and its direction is normal to the paper, coming out of it.
Thus,

τ = p × E

This torque will tend to align the dipole with the field E. When p is aligned with E, the torque is zero.

If the field is not uniform, In that case, the net force will evidently be non-zero.

The force depends on the orientation of p with respect to E.

### CONTINUOUS CHARGE DISTRIBUTION

A surface charge density σ at the area element by

σ=(ΔQ)/(ΔS)

σ= the surface charge density.

ΔQ= charge contain in area element.

The units for  σ are C/m^2 .

The linear charge density λ of a wire is defined by

λ = (ΔQ)/(Δl)

Δl = a small line element of wire

ΔQ is the charge contained in that line element.
The units for λ are C/m .

The volume charge density (sometimes simply called charge density)
is defined in a similar manner:

ρ=(ΔQ)/(ΔV)

ΔQ is the charge included in the small volume element ΔV that includes a large number of microscopic charged
constituents.

The units for ρ are C/m^3 .

Consider any general point P with position vector R . Electric field due to the charge ρΔV is given by Coulomb’s law :

ΔE=1/(4πε_0)(ρΔV)/r'^2hatr'

where  r′  = distance between the charge element and P

hatr′ = unit vector in the direction from the charge element to P.

### Gauss's Law

let us consider the total flux through a sphere of radius r, which encloses a point charge q at its centre. Divide the sphere into small area elements, as shown in Fig. below

Electric flux through a closed surface S

= q/ε_0

q = total charge enclosed by S.

Let us consider the total flux through a sphere of radius r, which encloses a point charge q at its centre. Divide the sphere into small area elements, as shown in

The flux through an area element ΔS is

Δφ=E.ΔS=q/(4πε_0r^2)r. Δ S

The unit vector hat r is along the radius vector from the centre to the area element. Now, since the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and hat r have the same direction. Therefore,

Δφ=q/(4πε_0r^2)ΔS

The total flux through the sphere is obtained by adding up flux through all the different area elements:

φ=sum_(allΔs )q/(4πε_0r^2 )ΔS

Since each area element of the sphere is at the same distance r from the charge,

φ=q/(4πε_0r^2 )sum_(allΔs )ΔS=q/(4πε_0r^2 )S

Now S, the total area of the sphere, equals 4πr^ 2 . Thus,

φ=(q/(4πε_0r^2 ))*(4πr^2) =q/ε_0

Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field E. The total
flux φ through the surface is φ = φ_1 + φ_2 + φ_3, where φ_1 and φ_2 represent the flux through the surfaces 1 and 2 of the cylinder and φ_3 is the flux through the curved cylindrical part of the closed surface. Now the normal to the surface 3 at every point is perpendicular to E, so by definition of flux, φ_3 = 0. Further, the outward normal to 2 is along E while the outward normal to 1 is opposite to E.

Therefore,
φ_1 = –E S_1, φ_2 = +E S_2

S_1 = S_2 = S

where S is the area of circular cross-section. Thus, the total flux is zero, as expected by Gauss’s law. Thus, whenever you find that the net electric flux through a closed surface is zero, we conclude that the total charge contained in the closed surface is zero.

\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT)
The great significance of Gauss’s law , is that it is true in general, and not only for the simple cases we have considered above. Let
us note some important points regarding this law:

color{blue}{(i)} Gauss’s law is true for any closed surface, no matter what its shape or size.

color{blue}{(ii)} The term q on the right side of Gauss’s law , includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.

color{blue}{(iii)} When the surface is so chosen that there are some charges inside and some outside, the electric field is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law, however, represents only the total charge inside S.

color{blue}{(iv)} The surface that we choose for the application of Gauss’s law is called the Gaussian surface. You may choose any Gaussian surface and apply Gauss’s law. However, take care not to let the Gaussian surface pass through any discrete charge. This is because electric field due to a system of discrete charges is not well defined at the location of any charge .However, the Gaussian surface can pass through a continuous charge distribution.

color{blue}{(v)} Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. This is facilitated by the choice of a suitable Gaussian surface.

color{blue}{(vi)} Finally, Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.
Q 3280680517

The electric field components in Fig. 1.27 are
E_x = α x^(1//2), E_y = E_z = 0, in which α = 800 N//C m^(1//2). Calculate (a) the
flux through the cube, and (b) the charge within the cube. Assume
that a = 0.1 m.

Class Chapter 1 Example 11
Solution:

(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and ΔS is
± π//2. Therefore, the flux phi = E.ΔS is separately zero for each face
of the cube except the two shaded ones. Now the magnitude of
the electric field at the left face is

E_L = αx ^(1//2) = αa^(1//2)

(x = a at the left face).

The magnitude of electric field at the right face is

E_R = α x^(1//2) = α (2a)^(1//2)

(x = 2a at the right face).

The corresponding fluxes are

phi_L= E_L .ΔS = ΔS E_L ⋅ hat n_L =E_L ΔS cosθ = –E_L ΔS, since θ = 180°

= –E_L a^2

phi_R= E_R .ΔS = E_R ΔS cosθ = E_R ΔS, since θ = 0°

= E_R a^2

Net flux through the cube

= phi_R + phi_L = E_R a^2 – E_La^2 = a^2 (E_R – E_L) = αa^2 [(2a)^(1//2) – a^(1//2) ]

= αa^(5//2) ( sqrt2 – 1)

= 800 (0.1)^(5//2) ( sqrt2 – 1)

= 1.05 N m^2 C^(–1)

(b) We can use Gauss’s law to find the total charge q inside the cube.
We have phi = q//ε_0 or q = phi ε_0. Therefore,

q = 1.05 xx 8.854 xx 10^(–12) C = 9.27 xx 10^(–12) C.
Q 3200680518

An electric field is uniform, and in the positive x
direction for positive x, and uniform with the same magnitude but in
the negative x direction for negative x. It is given that E = 200 hat i N//C
for x > 0 and E = –200 hat i N//C for x < 0. A right circular cylinder of
length 20 cm and radius 5 cm has its centre at the origin and its axis
along the x-axis so that one face is at x = +10 cm and the other is at
x = –10 cm (Fig. 1.28). (a) What is the net outward flux through each
flat face? (b) What is the flux through the side of the cylinder?
(c) What is the net outward flux through the cylinder? (d) What is the
net charge inside the cylinder?
Class Chapter 1 Example 12
Solution:

(a) We can see from the figure that on the left face E and ΔS are
parallel. Therefore, the outward flux is

phi _L= E.ΔS = – 200 hat i .ΔS

= + 200 Δ S, since hat i ΔS = – ΔS

= + 200 xx π (0.05)^2 = + 1.57 N m^2 C^(–1)

On the right face, E and ΔS are parallel and therefore

phi_R = E.ΔS = + 1.57 N m^2 C^(–1).

(b) For any point on the side of the cylinder E is perpendicular to
ΔS and hence E.ΔS = 0. Therefore, the flux out of the side of the
cylinder is zero.
(c) Net outward flux through the cylinder
phi = 1.57 + 1.57 + 0 = 3.14 N m^2 C^(–1)

(d) The net charge within the cylinder can be found by using Gauss’s
law which gives

q = ε_0 phi

= 3.14 xx 8.854 xx 10^(–12) C

= 2.78 xx 10^(–11) C

### Application of Gauss's Law

color{blue}{1.} color{blue}bbul{" Field due to an infinitely long straight uniformly charged wire:"}

Imagine a cylindrical Gaussian surface .Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero.

At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. The surface are of the curved part is 2πrl, where l is the length of the cylinder. Flux through the Gaussian surface

= flux through the curved cylindrical part of the surface

= E × 2πrl

The surface includes charge equal to λ_l . Gauss’s law then gives

 E × 2πrl = (λl)/(ε_0)

i.e.,  E =λ/(2πε_0r)

Vectorially, E at any point is given by:
 E =λ/(2πε_0r) hat n

where hat n is the radial unit vector in the plane normal to the wire passing through the point. E is directed outward if λ is positive and inward if λ is negative.

2.color{green}bbul{" Field due to a uniformly charged infinite plane shee"}

Let σ be the uniform surface charge density of an infinite plane sheet We can take the Gaussian surface to be a rectangular paralleleopiped of cross sectional area A

Therefore, flux E*ΔS through both the surfaces are equal and add up. Therefore the net flux through the Gaussian surface is 2 EA.

The charge enclosed by the closed surface is σA.

Therefore by Gauss’s law,

2 EA = σA/ε_0

or, E = σ/2ε_0

Vectorically,

E = σ/(2ε_0) hat n

where hat n is a unit vector normal to the plane and going away from it. E is directed away from the plate if σ is positive and toward the plate if σ is negative.

color{blue}{3.} color{blue}bbul{"Field due to a uniformly charged thin spherical shell"}

Let σ be the uniform surface charge density of a thin spherical shell of radius R. The field at any point P, outside or inside, can depend only on r .

color{blue}{(i)} Field outside the shell :

Consider a point P outside the shell with radius vector r. To calculate E at P, we take the Gaussian surface to be a sphere of radius r and with centre O, passing through P. All points on this sphere are equivalent relative to the given charged configuration. The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is along the radius vector at each point. Thus, E and ΔS at every point are parallel and the flux through each element is E ΔS. Summing over all ΔS, the flux through the Gaussian surface is

E × 4 π r ^2. The charge enclosed is σ × 4 π R^2. By Gauss’s law

E × 4 π r^ 2 = σ/ε_0(4πR^2)

Or, E = (σR^2)/(ε_0r^2) =q/(4πε_0r^2)

where  q = 4 π R^2 σ is the total charge on the spherical shell.

Vectorially, E =q/(4πε_0r^2) hat r

The electric field is directed outward if q > 0 and inward if q < 0. This, however, is exactly the field produced by a charge q placed at the centre O. Thus for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre.

color{blue}{(ii)} Field inside the shell :

The point P is inside the shell. The Gaussian surface is again a sphere through P centred at O.The flux through the Gaussian surface, calculated as before, is

 E × 4 π r^2.

However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives

E × 4 π r^2 = 0
i.e., E = 0" " (r < R )

that is, the field due to a uniformly charged thin shell is zero at all points inside the shell*. This important result is a direct consequence of Gauss’s law which follows from Coulomb’s law. The experimental verification of this result confirms the 1/r^2  dependence in Coulomb’s law.
Q 3210680519

An early model for an atom considered it to have a
positively charged point nucleus of charge Ze, surrounded by a
uniform density of negative charge up to a radius R. The atom as a
whole is neutral. For this model, what is the electric field at a distance
r  from the nucleus ?
Class Chapter 1 Example 13
Solution:

The charge distribution for this model of the atom is as
shown in Fig. 1.32. The total negative charge in the uniform spherical
charge distribution of radius R must be –Z e, since the atom (nucleus
of charge Z e + negative charge) is neutral. This immediately gives us
the negative charge density ρ, since we must have

 (4 pi R^3)/3 rho = 0 - ze

or  rho = -( 3 Ze)/(4 pi R^3)

To find the electric field E(r) at a point P which is a distance r away
from the nucleus, we use Gauss’s law. Because of the spherical
symmetry of the charge distribution, the magnitude of the electric
field E(r) depends only on the radial distance, no matter what the
direction of r. Its direction is along (or opposite to) the radius vector r
from the origin to the point P. The obvious Gaussian surface is a
spherical surface centred at the nucleus. We consider two situations,
namely, r < R and r > R.
(i) r < R : The electric flux phi enclosed by the spherical surface is
phi = E (r ) xx 4 π r^ 2
where E (r ) is the magnitude of the electric field at r. This is because

the field at any point on the spherical Gaussian surface has the
same direction as the normal to the surface there, and has the same
magnitude at all points on the surface.
The charge q enclosed by the Gaussian surface is the positive nuclear
charge and the negative charge within the sphere of radius r .

i.e  q = Ze + (4 pi r^3)/3 rho

Substituting for the charge density ρ obtained earlier, we have

 q = Ze - Ze r^3/R^3

Gauss’s law then gives,

 E(r) = (Ze)/(4 pi epsilon_0) (1/r^2 - r/R^3) : r < R

The electric field is directed radially outward.
(ii) r > R: In this case, the total charge enclosed by the Gaussian
spherical surface is zero since the atom is neutral. Thus, from Gauss’s
law,
E (r ) xx 4 π r^2 = 0 or E (r ) = 0 ; r > R
At r = R, both cases give the same result: E = 0.