Electric ar can be considered as an electric property linked with each suggest in the room where a charge is current in any type of form. An electric field is additionally described together the electrical force per unit charge.

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The formula of electric field is given as;

E = F /Q


E is the electrical field.F is a force.Q is the charge.

Electric areas are usually resulted in by differing magnetic areas or electric charges. Electrical field stamin is measured in the SI unit volt every meter (V/m).

The direction the the field is taken together the direction that the pressure which is exerted on the positive charge. The electrical field is radially outwards from optimistic charge and radially in towards negative point charge.



What is electrical Field?

The electrical field is defined mathematically together a vector ar that have the right to be associated with each allude in space, the pressure per unit charge exerted on a confident test fee at remainder at the point.

The electric field is created by the electrical charge or by time-varying magnetic fields. In the case of atom scale, the electric field is responsible because that the attractive forces in between the atom nucleus and electrons which host them together.

According come coulomb’s law, a fragment with electrical charge q1 at place x1 exerts a force on a particle with charge q0 at position x0 of,



r1,0 is the unit vector in the direction from suggest x1 to allude x0

ε0 is the electric continuous also recognized as pure permittivity of free space C2m-2N-1

When the charges q0 and also q1 have actually the very same sign then the force is positive, the direction is away from various other charges which method they repel every other. As soon as the charges have unlike signs then the pressure is an adverse and the particles lure each other.

Electric field is pressure per unit charge,


Electric ar can be uncovered easily by using Gauss regulation which claims that the full electric flux the end of a closed surface ar is same to the fee enclosed split by the permittivity.

Or full flux attached with a surface ar is 1/ ε0 time the fee enclosed by the closed surface.

∮E⃗⋅d⃗s=1ϵoq⋅\oint \vecE\cdot \vecds = \frac1\epsilon _oq\cdot∮E

⋅ds=ϵo​1​q⋅The electrical field can also be calculate by coulomb’s law but the Gauss law technique is easier. Besides, Gauss legislation is just a replica of the coulomb’s law. If we use Gauss theorem come a allude charge enclosed by a sphere, us will obtain to coulomb’s law.

How to discover Electric field Using Gauss law?

These incorporate a couple of steps.

Initially, us should find the spatial the contrary ( spherical, cylindrical, planar) of fee distribution.Next, we require to uncover a gaussian symmetry as exact same as that of symmetry of spatial arrangement.Find integral follow me the gaussian surface and also then uncover the flux.Find fee enclosed by Gaussian surface.Find the electric field of charge distribution. Electrical field as result of a point charge.

The electrical field is a vector field which is linked with the Coulomb pressure experienced by a test fee at each suggest in the room to the source charge. The magnitude and the direction of the electrical field can be identified by the Coulomb force F ~ above the test charge q. If the field is created by a confident charge, the electrical field will certainly be in radially outward direction and also if the ar is produced by an unfavorable charge, the electric field will be in radially inwards direction.

Let, a point charge Q be placed in a vacuum. Then if we introduce another allude charge q (test charge) at a street r native the charge Q.



Then the electrical field at point p because of the allude charge Q is provided by,

E⃗=F⃗Q\vecE = \frac\vecFQE

=QF​The direction the the electric field as result of a suggest charge Q is presented in the above figure. The size of the electric field is proportional to the size of E. If a test fee which is relatively larger is lugged within the area that the source charge Q that is bound to modify the original electric field due to source charge. A simple means to escape native this conflict is to use a an extremely negligible test charge q.

Then our definition for electric field modifies to,

E⃗=limq→0(F⃗Q)\vecE = \fraclimq\rightarrow 0(\frac\vecFQ)E

=q→0lim​(QF​)According come this definition,

The electric field at suggest P early out to suggest charge Q is,

E⃗=14πϵoQr2r^\vecE = \frac14\pi \epsilon o \fracQr^2 \hatrE


Electric Field as result of Line Charge

One applications of Gauss legislation is to uncover the electric field due to the charged particle. Electric field as result of line charge have the right to be found easily by making use of Gauss law. Consider,

A line charge is in the kind of a thin charged stick with straight charge density λ.



To uncover the electrical intensity at point P in ~ a perpendicular street r native the rod, because that that, allow us take into consideration a ideal circular closed cylinder the radius r and also length l through an infinitely lengthy line of charge as the axis.

The size of the electrical field intensity at every point on the curved surface of the Gaussian surface (cylinder) is the same since all points space at the exact same distance from line charge.

Therefore, the contribution of the curved surface ar of the cylinder towards electric flux,


2 Π rl is the curved surface area that the cylinder.

On the ends of the cylinder, the angle between the electric field and also its direction is 90o. Therefore these end of the cylinder will certainly not have any type of effect in the electrical flux.


ϕ E =q/ εo

Charge enclosed in cylinder=line charge thickness × length= λ l so follow to Gauss law,

ϕ E =q/ εo

E(2 Π rl)= λ l/ εo


E= λ /(2 Π εor)

Also Read: Faraday’s Law

Electric Field as result of Ring

We want to find the electrical field in ~ an axial point P because of a uniformly charged ring as presented in the figure,




The center of the ring is at suggest O.

The one of the circle makes an angle θ v line OP attracted from the center of the ring come the suggest P.

Now consider a little element from the ring together dq, currently calculate the field at point P because of this charge element,

We recognize that the electric field at suggest P is,

dE=dq4πϵoz2dE = \fracdq4\pi \epsilon oz^2dE=4πϵoz2dq​

The direction of the field is follow me AP. For this reason by resolving it, we gain one along the optimistic x-direction and also another along an adverse y-direction as shown over figure.

The x ingredient of the electrical field as result of charge aspect dq is,

dEx=dEcos(Θ)=dq4πϵoz2cos(Θ)dE_x = dEcos(\Theta) = \fracdq4\pi \epsilon oz^2 cos (\Theta )dEx​=dEcos(Θ)=4πϵoz2dq​cos(Θ)

The y ingredient of the electric field as result of charged aspect dq is,

dEx=dEsin(Θ)=dq4πϵoz2sin(Θ)dE_x = dEsin(\Theta) = \fracdq4\pi \epsilon oz^2 sin (\Theta )dEx​=dEsin(Θ)=4πϵoz2dq​sin(Θ)

To calculate the full electric ar at suggest P as result of charge ring we require to incorporate dE over the ring,

But as result of axial symmetry, the y component will certainly vanish and x component will just exist,


E=∫dq4πϵoz2cos(Θ)E = \int \fracdq4\pi \epsilon oz^2 cos (\Theta )E=∫4πϵoz2dq​cos(Θ)


E=cosΘ4πϵoz2∫dqE = \fraccos\Theta 4\pi \epsilon oz^2\int dqE=4πϵoz2cosΘ​∫dq


E=QcosΘ4πϵoz2E = \fracQcos\Theta 4\pi \epsilon oz^2E=4πϵoz2QcosΘ​


E=Qr4πϵo(a2+r2)32E = \fracQr 4\pi \epsilon o\left ( a^2 + r^2 \right )\frac32E=4πϵo(a2+r2)23​Qr​


E⃗=Qr4πϵo(a2+r2)32r^\vecE = \fracQr 4\pi \epsilon o\left ( a^2 + r^2 \right )\frac32\hatrE

=4πϵo(a2+r2)23​Qr​r^The direction of the field is follow me OP.

Electric field due to continuous Charge Distribution

Here we require to think about that the charges are spread continuously over a length or a surface ar or a volume.

If we want to find the electrical field v a surface ar which tote charges continuously over the surface, the is not possible to find the electrical field due to each fee constituent. Therefore in order to deal with this trouble we take into consideration an elementary area and also integrate it.

If the full charge brought by an area facet is equal to Δ Q, then the charge thickness of the element is,

σ=ΔQΔs\sigma = \frac\Delta Q\Delta sσ=ΔsΔQ​

σ is C/m2

Similarly, in the situation of charge distribution along the line segment of size Δl, the linear charge thickness is,

λ=ΔQΔl\lambda = \frac\Delta Q\Delta lλ=ΔlΔQ​

λ is C/m

Similarly, in case of fee distributed along with a volume element Δv, the volume thickness can be provided by,

ρ=ΔQΔV\rho = \frac\Delta Q\Delta Vρ=ΔVΔQ​

ρ is C/m3

Now allow us take into consideration a instance of the continuous charge distribution we will calculate the electric field because of this fee at allude P. Right here we have to divide the body right into different facets and then take into consideration one facet of volume Δv whose charge thickness is ρ. Climate let the street of the volume facet from allude P is given as r. Then fee in the volume aspect is ρ Δv. Then the electric field is,

ΔE=14πϵoρΔVr2r^\Delta E = \frac14\pi \epsilon o \frac\rho\Delta Vr^2\hatrΔE=4πϵo1​r2ρΔV​r^

Electric Field because of a Uniformly fee Sphere

Let σ be the uniform surface ar charge thickness of round of radius R.



Let us uncover out electric field intensity at a suggest P outside or inside the shell.

Field outside the Shell

We have to find the electrical field intensity in ~ a suggest P external the spherical covering such that, OP=r.

Here us take gaussian surface as a round of radius r.

Then the electrical field soot is the same at every point of gaussian surface directed radially outwards,

So, follow to Gauss’s theorem


So the is clear that electrical intensity at any point outside the spherical shell is such as if the whole charge is focused at the center of the shell.

Field at the surface of the Shell

Here us have, r = R


E = q / (4 Π R2 εO)

If σ C/m2 is the charge density on the shell,


q=4 Π R2. σ


E = (4 Π R2. σ ) / (4 Π R2. εO) = σ /ε0

Field inside the Shell

If the point P lies inside the spherical shell then the gaussian surface ar is a surface of a sphere of radius r.

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As there is no fee inside the spherical shell, the Gaussian surface encloses no charge. The is q = 0.