Electric Field

Electric field can be considered as an electric property associated with each point in the space where a charge is present in any form. An electric field is also described as the electric force per unit charge.

The formula of electric field is given as;

E = F /Q

Where,

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

Electric fields are usually caused by varying magnetic fields or electric charges. Electric field strength is measured in the SI unit volt per meter (V/m).

The direction of the field is taken as the direction of the force which is exerted on the positive charge. The electric field is radially outwards from positive charge and radially in towards negative point charge.

Direction of electric field

What is Electric Field?

The electric field is defined mathematically as a vector field which can be associated with each point in space, the force per unit charge exerted on a positive test charge at rest at that point.

Electric field is generated by the electric charge or by time-varying magnetic fields. In case of atomic scale, the electric field is responsible for the attractive forces between the atomic nucleus and electrons which hold then together.

According to coulomb’s law, a particle with electric charge q1 at position x1 exerts a force on a particle with charge q0 at position x0 of,

Coulomb's law

Where,

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

ε0 is the electric constant also known as absolute permittivity of free space C2m-2N-1

When the charges q0 and q1 have the same sign then the force is positive, the direction is away from other charges which means they repel each other. When the charges have unlike signs then the force is negative and the particles attract each other.

Electric field is force per unit charge,

Electric field formula

Electric field can be found easily by using Gauss law which states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

Or total flux linked with a surface is 1/ ε 0 times the charge enclosed by the closed surface.

Eds=1ϵoq\oint \vec{E}\cdot \vec{d}s = \frac{1}{\epsilon _{o}}q\cdot

The electric field can also be calculated by coulomb’s law but the Gauss law method is easier. Besides, Gauss law is just a replica of the coulomb’s law. If we apply Gauss theorem to a point charge enclosed by a sphere, we will get to coulomb’s law.

How to Find Electric Field Using Gauss law?

These include a few steps.

  1. Initially, we should find the spatial symmetry ( spherical, cylindrical, planar) of charge distribution.
  2. Next, we need to find a gaussian symmetry as same as that of symmetry of spatial arrangement.
  3. Find integral of ϕ sE along the gaussian surface and then find the flux.
  4. Find charge enclosed by Gaussian surface.
  5. Find the electric field of charge distribution. Electric field due to a point charge.

The electric field is a vector field which is associated with the Coulomb force experienced by a test charge at each point in the space to the source charge. The magnitude and the direction of the electric field can be determined by the Coulomb force F on the test charge q. If the field is created by a positive charge, the electric field will be in radially outward direction and if the field is created by negative charge, the electric field will be in radially inwards direction.

Let, a point charge Q placed in a vacuum. Then if we introduce another point charge q (test charge) at a distance r from the charge Q.

Electric field at point p due to the point charge Q

Then the electric field at point p due to the point charge Q is given by,

E=FQ\vec{E} = \frac{\vec{F}}{Q}

The direction of the electric field due to a point charge Q is shown in the above figure. The magnitude of the electric field is proportional to the length of E. If a test charge which is relatively larger is brought within the area of the source charge Q it is bound to modify the original electric field due to source charge. A simple way to escape from this conflict is to use a very negligible test charge q.

Then our definition for electric field modifies to,

E=limq0(FQ)\vec{E} = \frac{lim}{q\rightarrow 0}(\frac{\vec{F}}{Q})

According to this definition,

The electric field at point P due to point charge Q is,

E=14πϵoQr2r^\vec{E} = \frac{1}{4\pi \epsilon o} \frac{Q}{r^{2}} \hat{r}

Electric Field Due to Line Charge

One application of Gauss law is to find the electric field due to the charged particle. Electric field due to line charge can be found easily by using Gauss law. Consider,

A line charge is in the form of a thin charged rod with linear charge density λ.

Electric Field Due to Line Charge

To find the electric intensity at point P at a perpendicular distance r from the rod, For that, let us consider a right circular closed cylinder of radius r and length l with an infinitely long line of charge as its axis.

The magnitude of the electric field intensity at every point on the curved surface of the Gaussian surface (cylinder) is the same since all points are at the same distance from line charge.

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

Curved surface area of the cylinder

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

On the ends of the cylinder, the angle between the electric field and its direction is 90o. So these ends of the cylinder will not have any effect in the electric flux.

Therefore,

ϕ E =q/ εo

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

ϕ E =q/ εo

E(2 Π rl)= λ l/ εo

Or,

E= λ /(2 Π εor)

Also Read: Faraday’s Law

Electric Field Due to Ring

We want to find the electric field at an axial point P due to a uniformly charged ring as shown in the figure,

 

Electric Field Due to Ring

The centre of the ring is at point O.

The circumference of the circle makes an angle θ with line OP drawn from the centre of the ring to the point P.

Now consider a small element from the ring as dq, Now calculate the field at point P due to this charge element,

We know that the electric field at point P is,

dE=dq4πϵoz2dE = \frac{dq}{4\pi \epsilon oz^{2}}

The direction of the field is along AP. So by resolving it, we get one along the positive x-direction and another along negative y-direction as shown above figure.

The x component of the electric field due to charge element dq is,

dEx=dEcos(Θ)=dq4πϵoz2cos(Θ)dE_{x} = dEcos(\Theta) = \frac{dq}{4\pi \epsilon oz^{2}} cos (\Theta )

The y component of the electric field due to charged element dq is,

dEx=dEsin(Θ)=dq4πϵoz2sin(Θ)dE_{x} = dEsin(\Theta) = \frac{dq}{4\pi \epsilon oz^{2}} sin (\Theta )

To calculate the total electric field at point P due to charge ring we need to integrate dE over the ring,

But due to axial symmetry, the y component will vanish and x component will only exist,

Then,

E=dq4πϵoz2cos(Θ)E = \int \frac{dq}{4\pi \epsilon oz^{2}} cos (\Theta ) E=cosΘ4πϵoz2dqE = \frac{cos\Theta }{4\pi \epsilon oz^{2}}\int dq E=QcosΘ4πϵoz2E = \frac{Qcos\Theta }{4\pi \epsilon oz^{2}} E=Qr4πϵo(a2+r2)32E = \frac{Qr }{4\pi \epsilon o\left ( a^{2} + r^{2} \right )\frac{3}{2}}

Or

E=Qr4πϵo(a2+r2)32r^\vec{E} = \frac{Qr }{4\pi \epsilon o\left ( a^{2} + r^{2} \right )\frac{3}{2}}\hat{r}

The direction of the field is along OP.

Electric field due to Continuous Charge Distribution

Here we need to consider that the charges are distributed continuously over a length or a surface or a volume.

If we want to find the electric field with a surface which carries charges continuously over the surface, it is not possible to find the electric field due to each charged constituent. So in order to solve this problem we consider an elementary area and integrate it.

If the total charge carried by an area element is equal to Δ Q, then the charge density of the element is,

σ=ΔQΔs\sigma = \frac{\Delta Q}{\Delta s}

σ is C/m2

Similarly, in the case of charge distribution along the line segment of length Δl, the linear charge density is,

λ=ΔQΔt\lambda = \frac{\Delta Q}{\Delta t}

λ is C/m

Similarly, in case of charge distributed along with a volume element Δv, the volume density can be given by,

ρ=ΔQΔV\rho = \frac{\Delta Q}{\Delta V}

ρ is C/m3

Now let us consider a case of the continuous charge distribution we will calculate the electric field due to this charge at point P. Here we need to divide the body into different elements and

then consider one element of volume Δv whose charge density is ρ. Then let the distance of the volume element from point P is given as r. Then charge in the volume element is ρ Δv. Then the electric field is,

ΔE=14πϵoρΔVr2r^\Delta E = \frac{1}{4\pi \epsilon o} \frac{\rho\Delta V}{r^{2}}\hat{r}

Electric Field due to a Uniformly Charged Sphere

Let σ be the uniform surface charge density of sphere of radius R.

Electric Field due to a Uniformly Charged Sphere

Let us find out electric field intensity at a point P outside or inside the shell.

Field Outside the Shell

We have to find the electric field intensity at a point P outside the spherical shell such that, OP=r.

Here we take gaussian surface as a sphere of radius r.

Then the electric field intensity is the same at every point of gaussian surface directed radially outwards,

So, according to Gauss’s theorem

So it is clear that electric intensity at any point outside the spherical shell is such as if the entire charge is concentrated at the centre of the shell.

Field at the Surface of the Shell

Here we have, r = R

Therefore,

E = q / (4 Π R2 εO)

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

Then,

q=4 Π R2. σ

Therefore,

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 is a surface of a sphere of radius r.

Field Inside the Shell

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

Therefore,

E = 0

Hence the field inside the spherical shell is always zero.

BOOK

Free Class