The conductor possesses the efficiency of the free electrons that are drifting when the external voltage is supplied to the conductor and it starts conducting due to the mobility of these free electrons.
The electric field of a conductor is a result of the conductivity of the charges present on the per unit surface area of the conducting material and is given by the relation E= Q/ε0
Electric Field Inside a Conductor
The electric field inside a conductor is always zero. Inside the conductor, all the charges exert electrostatic forces on each other, and hence the net electric force on any charge is the sum of all the charges constituting inside the conductor. Moreover, all the charges are at the static equilibrium state.
In its static state, there is no charge present within or on the surface of the conductor and hence the electric field is zero. The charge carriers are distributed in such a way inside the conductor so that the electric field inside the conductor is zero everywhere. Hence, the electric field inside a conductor is zero.
Electric Field Outside a Conductor
The charged particles always settle on the surface of the conductor hence the electric field inside a conductor is zero.
There is a possibility that the charges can move either perpendicular or parallel to the surface of the conductor. But it is obvious that the charges cannot move outside the conductor and hence the electric field is non-zero in the direction perpendicular to the conductor, whereas the charges run parallel along the surface of the conductor hence the electric field is equal to zero.
Hence, the electric field outside the conductor is E=σ/ε0 and remains perpendicular to the surface of the conductor.
Electric Field at the Surface of a Charged Conductor
At the surface of a charged conductor, the electric field is the same as that present on the surface of the charged conductor and is constant at every point on and at the surface of the conductor.
If there are charges present at the surface then the electric field is a non-zero component along the surface because of the mobility of the free charges in presence of the electric field. That is why there will be some surface charge density per unit area of the conductor thus defining the electric field at the surface.
The electric flux through the surface of a charged conductor is given by Gauss Law
Φ =E.dA
On integrating
Φ=EA
The electric field due to the charged particle q is E=q/4πε0 r2
Substituting this in the above equation
E=q/4πε0 r2 (A)
Consider an electric flux passing through a small element of Gaussian surface which is nearly spherical, hence
Φ=q/4πε0 r2 (4πr2)
Therefore we get
EA =q/ε0
E=q/ε0A
The charge density is the total number of charges present per unit surface area of the conductor which is given by
σ =q/A
Hence we get
E=σε0
This is the electric field present at the surface of the charged conductor.
Electric Field Inside Cavity of a Conductor
Normally, the charge carriers inhabit the surface area of the conductor hence inside a cavity of a conductor the electric field will be zero.
In case the charge is placed within the cavity of a conductor then there will be conductivity at the cavity due to the presence of the surface charge density and hence the electric field will be equal to Σε0
But this is rarely possible. Moreover, there is electrostatic shielding within the conductor because of the density of the molecules and the potential difference between any two points in the cavity will be always zero therefore the electric field inside the cavity of a conductor is zero.
Electric Field Near a Charged Plane Conductor
Consider a charged plane sheet conductor having a surface charge density σ. Consider a small Gaussian surface dA on the plane conductor.
The electric flux passes through the plane has two surfaces hence the electric flux through both the surfaces adds up and we get,
Φ =2EA=q/ε0
The charge density is the ratio of charge per unit area of the charged plane conductor, therefore,
q=σA
2EA=σA/ε0
2E=σ/ε0
Hence the electric field through a charged plane conductor is
E=σ/2ε0
Electric Field on Surface of Conductor
Consider a small surface of a conducting material S. Let dA be a small element of a Gaussian surface and σ be the surface charge density of the surface.
By Gauss Law, the electric field through this element is
Φ =E.dA
=q/4πε0 r2 dA
The area of the small element be spherical in shape and therefore,
E.A=q/4πε0 r2* 4πr2
EA=q/ε0
Hence, E=q/ε0 (A)
This equation gives the electric field on the surface of the conductor.
Electric Field of a Long Straight Conductor
Consider a long straight current carrying conductor like a wire or a cylinder of length ‘l’ and a radius ‘r’. The surface charge density of the conductor is +σ. The direction of the electric field is shown in the figure below.
The electric flux through this wire is
Φ =EA
The surface area of the cylindrical Gaussian surface is A=2πrl and Φ=q/ε0.
So, we get,
q/ε0=E.2πrl
E=q/2πε0r
The charge per unit length of the cylindrical wire is denoted by λ
λ =q/I
Hence,
E=λ/2πε0r
This is the electric field produced on a long straight conductor.
Electric Field of a Spherical Conductor
We have discussed previously in this article that in a static state or in presence of the electric energy in the conductor, the electric field within the conductor is zero.
The charge carriers settle themselves on the surface of the conductor or at the surface of the conductor. For an electric field to exist in a spherical conductor there should be free electrons in the mobile state in response to the electric flux lines running through the conductor, but this doesn’t happen as no charges are present in the interior of the conductor.
Electric Field of a Charged Spherical Conductor
Suppose we have a spherical shaped conductor of radius ‘r’, the charge density on the surface of the spherical conductor is σ.
Let P be any point outside the spherical shell at a distance ‘R’ from the center of the sphere. The electric flux passing through a point P is
Φ =EA
EA=σA/ε0r
We are interested to find the electric field on the spherical shell of radius ‘R’ on which the point P lies. The area of the charged spherical shell is 4πr2
E.4πr2=σ/ε04πr2
E=σ/ε0r2R2
The electric field due to a charge particle at a distance R is
E=q/4πε0R2
Substituting this in the above equation we get
q/4πε0R2=σ/ε0r2R2
q/4π=σr2
We found that the charge on the surface of the spherical shell is
q=4πσr2
We know that if q>0 that is if the charge is positive then the direction of the electric field will be pointing outward and if q<0 that is the charge carrier is negative then the direction of the electric field is inward.
Now let us find out what is the electric field inside the spherical shell.
Assume the same spherical shell of radius ‘R’, but now the point P lies inside the spherical shell at a radius ‘r’.
There is no conductivity in the spherical shell and hence there is no electric flux through the interior of the shell. Hence, Φ =EA=E. 4π r2=0.
Electric Field of a Parallel Plate Conductor
Consider two parallel conducting plates each of length ‘l’ separated by the distance ‘d’. The electric current is passed through the plates and the surface charge density of the two plates is +σ and –σ respectively. The surface charge density of one plate is positive due to the positive charge carrier and that of another is negative due to the negative charge carriers.
The flux through the plate having positive charge isΦ =EA=q/ε0
Here we have two surfaces of the plate hence, Area=2A
The surface charge density is the ratio of charge per unit area, therefore q=σA.
Using this in the above equation we can write
E.2A=σ/ε0
E=σ/2ε0
This is the electric field due to the positive plate of the capacitor. The electric field due to the negatively charged plate of the capacitor is
E= – σ/2ε0
In the outer region of the capacitor, the total electric field due to both the capacitor plates is
E=σ/2ε0– σ/2ε0=0
Consider a point P between the two parallel plates. The electric field is in a direction from the positive to a negative plate which is opposite to the electric flux of the negative plate, thus adding up the electric field.
E=σ/2ε0+σ/2ε0
E=σ/ε0
This equation gives the electric field at any point between the two parallel plate conductors.
How to Find Electric Field of a Conductor?
The electric field of a conductor can be found by applying the Gauss Law which gives the resultant electric field due to the distribution of all the electric charges.
By knowing the charge density per unit area of the conductor, the total area of the conductor, the electric flux, and the permittivity of the material we can calculate the electric field of a conductor.
What is the electric field of a spherical conductor of radius 5.6 cm carrying a charge of -3C?
Given: q=-3C
r=5.6 cm=0.056m
The electric flux through the conductor is
Φ =q/ε0
=-3/8.85*10-12=33.9*1010
The area of the spherical shell is
A=4π r2
=4π* 0.056
=0.7 m2
The electric field of a spherical conductor is
E=Φ/A
=33.9*1010/0.7
=48.43*1010V/m
Hence, the electric field passing through the spherical conductor is 48.43*1010V/m.
Is the Electric Field Inside a Conductor is Zero?
Indeed! The electric field inside a conductor is always zero as all the charge carries lies on the surface area of the conductor.
According to the Gauss law, the electric flux through the conductor is 1/ε0 time the total charge of the conductor, but inside a conductor, there is no transportation of electric flux.
Why Should Electrostatic Field be Zero Inside a Conductor?
The charge carriers all reside on the surface of the conductor and then the electric flux line runs on the surface of the conducting material.
In a static condition as well as in presence of the electric source, the electrostatic force which is generated due to the migration of the charge is absent inside a conductor as there is no availability of the free charge within the conductor.
Electric Field of a Parallel Plate Capacitor
A capacitor stores the electric charge with it even after disconnecting it from the power source. The capacitance of the capacitor is a ratio of the charge per unit voltage which is formulated as
C=Q/V
Where C is a capacitance
Q is a charge stored by the capacitor
V is a potential difference between the two plates of the capacitor
The capacitor has two plates. On passing the electric current through the capacitor, one plate behaves as an anode and another as a cathode.
The electric flux through the capacitor is simply the potential difference between the plates per unit distance separating the two plates.
E=V/d
The electric field due to the charge plate we have found above as
E=σ/ε0
because Σ =Q/A which is a surface charge density
E=Q/ε0
The potential difference between the two plates is the entire potential difference from distance 0 to d.
Substituting the value of electric field for the capacitor plate we have
V=Qd/ε0 A
Hence, the capacitance of the plates is
C=ε0 A/d
Frequently Asked Questions
What is the electric field at a point situated at a distance of 15cm from the center of a spherical shell of radius 7cm having a surface charge density of 50C/m2?
Given: r=7cm=0.07m
R=15cm=0.15m
σ= 50C/m2
The electric field at a point outside the spherical shell is
E=σ/ε0 r2R2
=50/(8.85*10-12*0.072*0.152)
=50/(8.85* 10-12* 4.9* 10-3*22.5* 10-3)
=1.23*1012V/m
The electric field at a point 15cm away from the center of the spherical shell is 1.23*101012V/m.
What is the electric field due to a square sheet having a charge of +2C and a length of 3cm?
Given: Q=+2C
l=3cm=0.03m
The area of a square sheet is A=l2=0.03m2=9*10-4m2
The surface charge density on the sheet is
Σ =Q/A
=2.2*103C/m2
The electric field due to square sheet is
E=Σ/2ε0
=2.2*103/2*8.85*10-12
=12.42*1013V/m
The electric filed due to a square sheet is12.42*1013V/m.
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- Electric field in capacitor
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Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess.
