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Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad.
MAXWELL’s
EQUATIONS:
1. Maxwell’s equations, explanation.
2. Gauss law for electrostatics
3. Gauss law in differential form.
Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad.
There are four Maxwell’s equations. The equations and explanations are as follows.
As such, there is no “sink” or “source” for B (magnetic lines of force cannot be absorb or emerge) - the field
lines have no beginning and no end. All magnetic “charge” is found in a dipole, with a North and a South.
2) ∇ × E = −
dB
dt
The above equation is given by the negative of the
change in the magnetic field in time and this equation is
called as the curl of the electric field.
In other words, if the magnetic field isn’t changing,
electric field lines are straight. If it is changing, the
electric field “swirls” appropriately, depending on if the
field is increasing or decreasing. A changing magnetic
field can induce an electric field! (i.e. Faraday
induction). The negative sign is called Lenz’s law.
1) ∇ ⋅ E=ρ/ϵ0
This equation is called divergence of electric field and is
given by how much is strength of electric field at a given
point due to density of charge enclosed in a closed
surface. If there is more charge inside, the divergence is
greater. If it’s zero, the divergence is zero.
his is also called as the Gauss law of electric field, which
states that the divergence of the electric field is equal to
the charge density inside of a closed surface of interest,
multiplied by a constant.
3) ∇ ⋅ B = 0
The above equation says that there is no effective field for
the magnetic field like electric field. There are no
magnetic monopoles like electric monopoles. The
magnetic field lines start from one pole of magnet and
compulsory end at opposite of the magnet. i.e. there is
only dipole or the divergence of B is always zero.
Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad.
Gauss law for electrostatics derivation:
The Gauss law gives in site into study of the distribution of charges by knowing the electric field at a given
point and vice versa. The static charges may give out the electric flux or lines of force which create the
electric field strength or electric field intensity at given point.
The Guass law states that, electric flux through any closed surface, is equal to 1/ε times the total charge
enclosed by the surface. By convention, a positive electric charge generates a positive electric field. The law
was published posthumously in 1867 as part of a collection of work by the famous German mathematician
Carl Friedrich Gauss.
ФE=∫E.dS=q/ε
Where ε is the permittivity of the medium (for free space ε=ε0),
So ФE=∫E.dS=q/ε0
Where ∫E.dS is surface integral over the closed surface and q is the charge present in the closed surface S
The imaginary closed surface is called Gaussian Surface. If the surface encloses a continuous charge
distribution then q is replaced by the integral ∫ρdV, where ρ is the volume charge density.
Proof of gauss law of electrostatics:
Consider a source producing the electric field E is a point charge +q situated at a point O inside a volume
enclosed by an arbitrary closed surface S. let us consider a small area element ds around a point P on the
surface where the electric field produced by the charge +q is E. if E is along OP and area vector dS is along
the outward drawn normal to the area element dS, (Try to make the diagram yourself),
4) ∇ × B = μ0J + μ0ϵ0
dE
dt
It is known that increase in the magnetic field at a point
increases the current in density. In addition to with this, an
electric field will be generated due to the magnetic field.
The magnetic field induced around a closed loop is proportional to the
electric current plus displacement current (rate of change of electric field)
that the loop encloses. The curl of B or the “swirliness of B” is
equal to the Current Density (the amount of current per unit
volume) plus any change in the electric field. This second part
is often called the “displacement current, since it helps with
dealing the capacitors.
Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad.
Then the electric flux dФ through the area element dS is given by,
dФ=E.dS=EdS cosθ
Where θ is the angle between E and dS and it is zero degree, therefore (In fig. shown dA=dS)
dФ = EdS (1)
Electric intensity “E” at a point on a elemental area dS (dA in fig), due to the point charge ‘q’ situated at a
distance R is given by,
E =
q
4πϵoR2 =
q
dS×εo
Or E. dS =
q
εo
--------------------------2
From equation 1 and 2 we have
dФ=E.dS= q /εo ----------------------------- 3
Hence , total electric flux Ф through the entire closed surface S would be
Or Ф=∫E.dS= Q/εo ------------------------------- 4
Where Q = Σ(q1+q2+q3-q4-q5-q6…….)
Equation (4) represents Gauss law for electrostatics for a single point charge. If the source producing the
electric field has more than one point charges such as +q1,+q2,+q3,-q4.-q5,-q6…… etc,then the total flux due to
all of them would be the algebric sum of all the fluxes as,
q
Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad.
Ф=∫E.dS=1/ε0
Or Ф=∫E.dS=∑q/ε0
Gauss law in differential form derivation:
STATEMENT:-Differential form of Gauss law states that the divergence of electric field E at any point in
space is equal to 1/ε0 times the volume charge density,ρ, at that point.
Del.E= ρ/εo
Where ρ is the volume charge density (charge per unit volume) and ε0the permittivity of free space. It is one of
the Maxwell’s equations.
Derivation or Proof .Consider a region of continuous charge distribution with varying volume density of
charge ρ(charge per unit volume).In this region, consider a volume V enclosed by the surface S.if dV is an
infinitesimal small volume element enclosed by the surface dS, then according to Gauss’s law for a continuous
charge distribution
∫E.dS=1/ε0∫ρdV (1) Charge density ρ = charge/Volume = q/dV
According to Gauss-divergence theorem (q = ρ x dV)
∫E.dS=∫(DivE) dV (2)
By comparing equations(1) and (2),we get
=∫(DivE) dV = 1/εo∫ρdV
∫(DivE - ρ/ε0) dV = 0
As the volume under consideration is arbitrary, therefore,
DivE- ρ/ε0=0 or DivE = ρ/ε0 or ∇. E =
ρ
εo

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Maxwell's equations and their derivations.

  • 1. Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad. MAXWELL’s EQUATIONS: 1. Maxwell’s equations, explanation. 2. Gauss law for electrostatics 3. Gauss law in differential form.
  • 2. Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad. There are four Maxwell’s equations. The equations and explanations are as follows. As such, there is no “sink” or “source” for B (magnetic lines of force cannot be absorb or emerge) - the field lines have no beginning and no end. All magnetic “charge” is found in a dipole, with a North and a South. 2) ∇ × E = − dB dt The above equation is given by the negative of the change in the magnetic field in time and this equation is called as the curl of the electric field. In other words, if the magnetic field isn’t changing, electric field lines are straight. If it is changing, the electric field “swirls” appropriately, depending on if the field is increasing or decreasing. A changing magnetic field can induce an electric field! (i.e. Faraday induction). The negative sign is called Lenz’s law. 1) ∇ ⋅ E=ρ/ϵ0 This equation is called divergence of electric field and is given by how much is strength of electric field at a given point due to density of charge enclosed in a closed surface. If there is more charge inside, the divergence is greater. If it’s zero, the divergence is zero. his is also called as the Gauss law of electric field, which states that the divergence of the electric field is equal to the charge density inside of a closed surface of interest, multiplied by a constant. 3) ∇ ⋅ B = 0 The above equation says that there is no effective field for the magnetic field like electric field. There are no magnetic monopoles like electric monopoles. The magnetic field lines start from one pole of magnet and compulsory end at opposite of the magnet. i.e. there is only dipole or the divergence of B is always zero.
  • 3. Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad. Gauss law for electrostatics derivation: The Gauss law gives in site into study of the distribution of charges by knowing the electric field at a given point and vice versa. The static charges may give out the electric flux or lines of force which create the electric field strength or electric field intensity at given point. The Guass law states that, electric flux through any closed surface, is equal to 1/ε times the total charge enclosed by the surface. By convention, a positive electric charge generates a positive electric field. The law was published posthumously in 1867 as part of a collection of work by the famous German mathematician Carl Friedrich Gauss. ФE=∫E.dS=q/ε Where ε is the permittivity of the medium (for free space ε=ε0), So ФE=∫E.dS=q/ε0 Where ∫E.dS is surface integral over the closed surface and q is the charge present in the closed surface S The imaginary closed surface is called Gaussian Surface. If the surface encloses a continuous charge distribution then q is replaced by the integral ∫ρdV, where ρ is the volume charge density. Proof of gauss law of electrostatics: Consider a source producing the electric field E is a point charge +q situated at a point O inside a volume enclosed by an arbitrary closed surface S. let us consider a small area element ds around a point P on the surface where the electric field produced by the charge +q is E. if E is along OP and area vector dS is along the outward drawn normal to the area element dS, (Try to make the diagram yourself), 4) ∇ × B = μ0J + μ0ϵ0 dE dt It is known that increase in the magnetic field at a point increases the current in density. In addition to with this, an electric field will be generated due to the magnetic field. The magnetic field induced around a closed loop is proportional to the electric current plus displacement current (rate of change of electric field) that the loop encloses. The curl of B or the “swirliness of B” is equal to the Current Density (the amount of current per unit volume) plus any change in the electric field. This second part is often called the “displacement current, since it helps with dealing the capacitors.
  • 4. Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad. Then the electric flux dФ through the area element dS is given by, dФ=E.dS=EdS cosθ Where θ is the angle between E and dS and it is zero degree, therefore (In fig. shown dA=dS) dФ = EdS (1) Electric intensity “E” at a point on a elemental area dS (dA in fig), due to the point charge ‘q’ situated at a distance R is given by, E = q 4πϵoR2 = q dS×εo Or E. dS = q εo --------------------------2 From equation 1 and 2 we have dФ=E.dS= q /εo ----------------------------- 3 Hence , total electric flux Ф through the entire closed surface S would be Or Ф=∫E.dS= Q/εo ------------------------------- 4 Where Q = Σ(q1+q2+q3-q4-q5-q6…….) Equation (4) represents Gauss law for electrostatics for a single point charge. If the source producing the electric field has more than one point charges such as +q1,+q2,+q3,-q4.-q5,-q6…… etc,then the total flux due to all of them would be the algebric sum of all the fluxes as, q
  • 5. Engineering physics study materials by Praveen N Vaidya, SDMCET, Dharwad. Ф=∫E.dS=1/ε0 Or Ф=∫E.dS=∑q/ε0 Gauss law in differential form derivation: STATEMENT:-Differential form of Gauss law states that the divergence of electric field E at any point in space is equal to 1/ε0 times the volume charge density,ρ, at that point. Del.E= ρ/εo Where ρ is the volume charge density (charge per unit volume) and ε0the permittivity of free space. It is one of the Maxwell’s equations. Derivation or Proof .Consider a region of continuous charge distribution with varying volume density of charge ρ(charge per unit volume).In this region, consider a volume V enclosed by the surface S.if dV is an infinitesimal small volume element enclosed by the surface dS, then according to Gauss’s law for a continuous charge distribution ∫E.dS=1/ε0∫ρdV (1) Charge density ρ = charge/Volume = q/dV According to Gauss-divergence theorem (q = ρ x dV) ∫E.dS=∫(DivE) dV (2) By comparing equations(1) and (2),we get =∫(DivE) dV = 1/εo∫ρdV ∫(DivE - ρ/ε0) dV = 0 As the volume under consideration is arbitrary, therefore, DivE- ρ/ε0=0 or DivE = ρ/ε0 or ∇. E = ρ εo