Class 11 Physics Electric Field Notes

ELECTRICITY
CLASS 11 • PHYSICS

Electric Field

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Relative Permittivity

The permittivity of any medium with respect to the permittivity of free space (or vacuum) is called the relative permittivity of that medium. It is denoted by εr and is given by:

εr = permittivity of medium permittivity of free space
∴ εr = ε/ε0

The relative permittivity of the medium is also known as the dielectric constant of the medium. It is denoted by K.

K = εr = ε/ε0

Thus, for SI system and in a medium other than air:

F = 1 4πε0εr × q1q2 r2

OR

F = 1 4πε0K × q1q2 r2

Permittivity

The ability of a medium to pass the electric charge through that medium is called permittivity of that medium.

Electric Field

The space around the electric charge where the electric force of attraction or repulsion exists is called electric field.

Test Charge

The positive charge having unit magnitude is taken as test charge in electrostatics. It is denoted by q0.

q0 = +1 C

Electric Field Intensity

Electric field intensity at a point in an electric field is defined as the force experienced by a unit positive charge placed at that point.

If F is the force experienced by the unit positive test charge q0 at a point in an electric field, the electric field intensity is given by:

E = F/q0

E is a vector quantity and its SI unit is N/C.

Electric Field Intensity Due to a Point (Test) Charge

Let us consider a charge +q at point O in space and also consider a point P at distance r from O so that OP = r. If a test charge +q0 is placed at P, the force experienced by the test charge +q0 is given by:

Electric field intensity due to a point charge Positive charge q is at point O and positive test charge q0 is at point P, separated by distance r. Electric field points away from q. +q O +q₀ P r E
Electric field intensity due to a point charge.

For Vacuum

F = 1 4πε0 × qq0 r2 (i)

By definition, the magnitude of electric field intensity E at a point at distance r from charge +q is:

E = q 4πε0r2

For Other Medium

E = q 4πεr2

Electric Flux (φ)

The number of electric lines of force passing through a given surface when field is perpendicular to the direction of line of force is called electric flux.

Electric flux through a surface Electric field lines originating from a point pass through a rectangular surface of area A. Area (A) P
Flux through a surface of area A.

Mathematically, electric flux is defined as the product of electric field intensity and surface area when the field lines are parallel to the surface area vector.

Electric flux (φ) = EA

Where, E = electric field intensity and A = surface area.

In case the surface area vector is perpendicular to the field lines, θ = 90°, then:

φ = EA cosθ = EA cos90° = 0

There is no flux through the surface parallel to the field.

Gauss’s Theorem

It states that the total electric flux passing through a closed surface enclosing a charge is equal to 1/ε0 times the magnitude of net charge enclosed by the closed surface.
φ = q/ε0

Application of Gauss’s Theorem: Electric Field Due to a Charged Sphere

(i) At a Point Outside the Sphere

Let us consider a point P which is at distance r from the centre of a sphere of radius R at which the electric field intensity is to be determined. For this, draw a Gaussian surface through point P enclosing the charge +q, which is also a sphere of radius r.

Point outside a charged sphere A charged sphere of radius R is enclosed by a larger Gaussian sphere of radius r. Point P lies on the Gaussian surface and electric field E points radially outward. + + + + + + + O R r P E
Point P lying outside the charged sphere.
A = 4πr2 (i)

If E is the electric field intensity at point P, then electric flux passing through the Gaussian surface is:

φ = EA = E·4πr2 (ii)

Also, from Gauss’s theorem:

φ = q/ε0 (iii)

From equations (ii) and (iii):

E·4πr2 = q/ε0
∴ E = q 4πε0r2

Thus, this is the required expression for electric field intensity due to a charged sphere when the point lies outside the sphere.

(ii) At a Point on the Surface of Sphere

In this case Gaussian surface has radius equal to the charged sphere, i.e. r = R.

Point on the surface of a charged sphere Point P lies on the surface of a positively charged sphere of radius R. Electric field E points radially outward. + + + + + + + O r = R P E
Point P lying on the surface of the sphere.
A = 4πR2 (i)
φ = EA = E·4πR2 (ii)
φ = q/ε0 (iii)

From equations (ii) and (iii):

E·4πR2 = q/ε0
E = q 4πε0R2

This is the required expression for electric field intensity due to a charged sphere when the point lies on the surface of sphere.

(iii) When Point Lies Inside the Sphere

In this case Gaussian surface does not enclose any charge (i.e. q = 0). If E is the electric field intensity at point P, then the total electric flux passing through the Gaussian surface is given by:

Point inside a hollow charged sphere A smaller Gaussian surface of radius r lies inside a charged hollow sphere of radius R. Point P is on the inner Gaussian surface. + + + + + + + r P R
Point P lying inside the hollow charged sphere.
φ = EA = E·4πr2 (i)

Also, from Gauss’s theorem:

φ = q/ε0 (ii)

From equations (i) and (ii):

E·4πr2 = q/ε0
E = 0   [Since q = 0]

Hence, electric field intensity due to a charged sphere when the point lies inside the hollow sphere must be zero.

Surface Charge Density

The amount of electric charge per unit area of a charged surface of the conductor is called surface charge density. It is denoted by σ.

Surface charge density = Electric charge Surface area
∴ σ = q/A

Electric Field Due to a Charged Plane Conductor

Let us consider a charged plane conductor with uniform surface charge density σ. Let P be the point outside the charged plane conductor about which electric field intensity is to be determined. For this, draw a Gaussian surface of surface area A as shown in the figure.

Electric field due to a charged plane conductor A positively charged plane conductor is intersected by a pillbox Gaussian surface. Point P lies outside the conductor and electric field E is normal to the surface. + + + + + + + + + + + A P E
Charged plane conductor with Gaussian surface.

If E is the electric field intensity, then flux is:

φ = EA (i)

Also, the net charge q enclosed by the Gaussian surface is:

q = σA (ii)

From Gauss’s theorem:

φ = q/ε0 (iii)

Using equation (ii) in equation (iii):

φ = σA/ε0 (iv)

Comparing equations (i) and (iv):

EA = σA/ε0
∴ E = σ/ε0

This is the required expression for electric field intensity due to a charged plane conductor.

Field Outside a Charged Plane Conductor

Let us consider a charged plane conductor with uniform surface charge density σ. Let P be any point outside the charged plane conductor about which electric field intensity is to be determined. For this, draw a Gaussian surface of surface area A as shown in the figure.

Field outside a charged plane conductor A positively charged plane has a cylindrical Gaussian pillbox extending outside it. Its outer face has area A and point P lies inside the pillbox. + + + + + + + + P A σ
Field outside a charged plane conductor.
φ = EA (i)
q = σA (ii)
φ = q/ε0 (iii)

From equations (i), (ii) and (iii):

EA = σA/ε0
∴ E = σ/ε0

This is the required expression for electric field intensity outside a charged plane conductor.

Linear Charge Density (λ)

Charge per unit length of the conductor is called linear charge density.

λ = q/l

Electric Field Intensity Due to Linear Charge Density

Let us consider an infinitely long straight conductor of uniform linear charge density λ. Let P be any point about which electric field intensity is to be determined. For this, draw a Gaussian surface of length l and radius r as shown in the figure.

Infinite long conductor with cylindrical Gaussian surface An infinitely long charged straight conductor with linear charge density lambda lies along the axis of a cylindrical Gaussian surface of radius r and length l. Point P lies on the curved surface. + + + + + + + λ r P l
Infinite long charged conductor with cylindrical Gaussian surface.

If E is the electric field intensity, then flux is:

φ = EA = E·2πrl (i)

Also, net charge q enclosed by the Gaussian surface is:

q = λl (ii)

From Gauss’s theorem:

φ = q/ε0 = λl/ε0 (iii)

From equations (i) and (iii):

E·2πrl = λl/ε0
∴ E = λ 2πε0r

This is the required expression for electric field intensity due to linear charge density.

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