Electric Field Due To A Uniformly Charged Infinite Plane Sheet

Electric field intensity due to a uniformly charged infinite plane sheet can be calculated using Gauss law. In a uniformly charged plane sheet, electric charges are uniformly distributed over the entire surface of the sheet. The electric field produced due to charges at rest is perpendicular to the surface of the sheet. 

• Gauss' law is equivalent to Coulomb's law in that it represents the link between the electric field and the electric charge. 
• Gauss' law provides a comparable method for calculating electric intensity expressions.

The expression for the electric field intensity (E) due to a uniformly charged infinite plane sheet having surface charge density σ, is given as

\(E = \frac{\sigma}{2 \epsilon _0}\)

Where ϵo pronounced as epsilon naught is the absolute permittivity of free space and its value is 8.854 x 10^-12 C² N^-1 m^-2

Key Terms: Gauss Law, Electric Field, Electric charges, Electric flux, Charge density, Permittivity, Gaussian surface, Electric field lines.

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Electric Field

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The region or space around a charged body within which its influence can be felt by other small charges is known as an Electric field. 

• The electric field in a region is represented by imaginary lines, known as lines of force or electric field lines.
• Electric field lines are the straight or curved imaginary lines in a region such that the tangent at any point on the field line gives the direction of the electric field at that point in the region.

Properties of Electric Field Lines

Some properties of electric field lines are:

• Electric field lines begin from a positive charge and end on a negative charge.
• These are the imaginary lines and a tangent at a point on these lines gives the direction of the electric field at that point.
• Electric field lines do not form a closed loop.
• In a charge-free region, electric field lines can be taken to be continuous curves without any break.

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What is Gauss Law? 

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Gauss Law, often known as Gauss' flux theorem Gauss' theorem, is the law that specifies the link between the distribution of electric charges and the resulting electric field.

• Gauss' law states that the total amount of electric flux passing through any closed surface is proportionate to the enclosed electric charge. 
• It describes the electrical charge contained within or present within the enclosed closed surface.

According to Gauss Law, the total electric flux contained within a closed surface equals 1/ε₀ times the total electric charge enclosed by the closed surface. It is represented by

Φ = \(\oint_S \vec{E} . \vec{dS}  = \frac{Q}{\epsilon_o}\)

Where

• Φ is the total flux
• ε₀ is an electric constant, known as the absolute permittivity of free space
• Q is the total electric charge enclosed by the surface

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Electric Field Due To Uniformly Charged Infinite Plane Sheet

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Consider a thin infinite plane sheet of surface area A, on which charge Q is uniformly distributed over the surface of the sheet. Let σ be the surface charge density of the plane sheet.

Gaussian Surface for Uniformly Charged Infinite Plane Sheet

According to Gauss’s law

\(\phi = \oint_S \vec{E} . \vec{dA}  = \frac{Q}{\epsilon_o}= \frac{\sigma A}{\epsilon_o}\)…………….(1)

Where

• E is the electric field intensity at equidistant from the plane sheet
• ε₀ is the absolute permittivity of free space

Imagine a cylindrical Gaussian surface with an axis perpendicular to the plane of the sheet. This Gaussian surface is divided into three parts I, II, and III i.e. two end caps and the curved surface of the cylindrical Gaussian surface.

The net electric flux through the surface will be determined by integrating the product of electric field E and the area element dA, i.e.

\(\oint_S \vec{E} . \vec{dA} = \int_I \vec{E} . \vec{dA} + \int_{II} \vec{E} . \vec{dA} + \int_{III} \vec{E} . \vec{dA}\)

⇒ \(\oint_S \vec{E} . \vec{dA} = \int_I EdA cos \theta + \int_{II} EdA cos \theta + \int_{III} EdA cos \theta\)

For surface I and II, the angle between E and dA is 0o and for curved surface III, the angle between E and dA is 90. Therefore

\(\oint_S \vec{E} . \vec{dA} = \int_I EdA cos0 + \int_{II} EdA cos0 + \int_{III} EdA cos90\)

⇒ \(\oint_S \vec{E} . \vec{dA} = \int_I EdA + \int_{II} EdA+0\)

⇒ \(\oint_S \vec{E} . \vec{dA} = \int_I EdA + \int_{II} EdA\)

Using equation (1), we get

\(\phi = \int_I EdA + \int_{II} EdA\)

The electric field is uniform through the surface, therefore, we take E out of integration.

⇒ \(\phi = E \int_I dA + E \int_{II} dA\)

Also, ∫_I dA = ∫_II dA = A, surface area of the plane sheet

⇒ Φ = EA + EA = 2EA

From equation (1),

2EA = \( \frac{\sigma A}{\epsilon_o}\)

∴ \(E = \frac{\sigma}{2 \epsilon _0}\)

Above equation represents the expression for electric field intensity due to a uniformly charged infinite plane sheet.

The observation from the expression is:

• The electric field intensity is directly proportional to the surface charge density of the plane sheet.
• It is independent of the surface area of the sheet.
• Electric field intensity at a point is independent of the distance from the sheet.
• If surface charge density (σ) is positive then the direction of the electric field is outward and perpendicular to the plane of the sheet.
• If surface charge density (σ) is negative then the direction of the electric field is inward and perpendicular to the plane of the sheet.

Electric field intensity due to a uniformly charged infinite plane sheet, in vector form can be written as

\(E = \frac{\sigma}{2 \epsilon _0} \hat{n}\)

Where, n is a unit vector in a direction perpendicular to the plane of the sheet.

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Electric Field Due To Two Parallel Uniformly Charged Infinite Plane Sheets

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Consider two parallel infinite plane sheets A and B of uniform surface charge densities σ₁ and σ₂. The electric field intensity due to uniformly charged plane sheets can be calculated in three regions.

1. Region I: Outer region of plate A towards the -x axis.
2. Region II: Region between plate A and plate B
3. Region III: Outer region of plate B towards +x axis

Electric field intensity due to two plane sheets

We will take the following assumptions:

• Electric field pointing towards the positive x-axis will be taken as positive.
• Electric field pointing towards the negative x-axis will be taken as negative.

Electric field intensity due to a uniformly charged infinite plane sheet, in vector form is given by

\(E = \frac{\sigma}{2 \epsilon _0} \hat{n}\)

Therefore, Electric field intensity due to sheets A is given by

\(E_1 = \frac{\sigma_1}{2 \epsilon _0} \hat{i}\)

And, Electric field intensity due to sheets B is given by

\(E_2 = \frac{\sigma_2}{2 \epsilon _0} \hat{i}\)

Now, we will calculate the net electric in different regions.

In Region I:

The net electric field is given by

E_I = – E₁ – E₂

⇒ E_I = – \(\frac{\sigma_1}{2 \epsilon _0} \hat{i} - \frac{\sigma_2}{2 \epsilon _0} \hat{i}\)

⇒ E_I = – \(( \frac{1}{2 \epsilon _o})\) (σ₁ + σ₂)\(\hat{i}\)

In Region II:

The net electric field is given by

E_II = – E₁ – E₂

⇒ E_II = \(\frac{\sigma_1}{2 \epsilon _0} \hat{i} - \frac{\sigma_2}{2 \epsilon _0} \hat{i}\)

⇒ E_II = \(( \frac{1}{2 \epsilon _o})\)(σ₁ – σ₂)\(\hat{i}\)

In Region III:

The net electric field is given by

E_III = – E₁ – E₂

⇒ E_III = – \(\frac{\sigma_1}{2 \epsilon _0} \hat{i} - \frac{\sigma_2}{2 \epsilon _0} \hat{i}\)

⇒ E_III = – \(( \frac{1}{2 \epsilon _o})\) (σ₁ + σ₂)\(\hat{i}\)

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Things to Remember

• Electric field is the 3D space around a charged body where a small charge experienced a force.
• Gauss Law states that the electric flux across any closed surface is proportional to the net electric charge enclosed by the surface.
• Electric Field due to uniformly Charged Infinite Plane Sheet is given by E = σ/2ε_o
• The symbol ϵo is the absolute permittivity of free space and its value is 8.854 x 10^-12 C² N^-1 m^-2
• Gaussian surface is an imaginary surface around a charged body such that electric field intensity is constant at every of it.