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Electric Field due to Infinite Plane for JEE

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Last updated date: 28th May 2024
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Electric Field - Brief Introduction

An electric field can be explained to be an invisible field around the charged particles where the electrical force of attraction or repulsion can be experienced by the charged particles. It is also defined as electrical force per unit charge. The SI unit of measurement of electric field is Volt/metre. The formula to determine the electric field is given as 

$E=\dfrac{F}{q}$

Where E is the electric field, F is the electric force and q is the charge.


Since the electric field is an invisible field, we use electric field lines to visualise the electric fields. This concept was introduced by Michael Faraday. The electric field lines are drawn in a tangential direction to the net electric field at a point. 


Some basic properties of Electric field lines are listed below.

  1. Electric field lines are always drawn perpendicular to the charge surface.

  2. Electric field lines do not intersect each other.

  3. Electric field lines start from a positive charge and end at a negative charge.

  4. When there is a single charge, if the charge is negative, then the electric field lines start from infinity and end at the charge; and if the charge is positive, then the electric field lines start from the charge and end at infinity.

  5. Electric field lines and the magnitude of a charge, these are directly proportional to each other.

 

Gauss’ Law

In electrostatics, we study about the electric charges at rest. The electric charges form an electric field around them, thus affecting the properties in the environment surrounding the charges. The electric field is stated to be a property of a charged system. A mathematician named Karl Friedrich Gauss (1777-1855), formulated a law known as Gauss’ law. This law explains the connection between electric fields and the electric charges. Gauss’ law helps to determine the intensity of electric fields due to various charged surfaces.


The statement of Gauss’ Law is that “The total flux contained within a closed surface equals 1/ε0 times the total electric charge enclosed by the closed surface.” According to Gauss' theorem, we know that ‘Within a closed surface, the net electric charge is  proportional to total electric flux enclosed by the surface. Thus, if φ represents the total electric flux and if the electric permittivity constant is ε0, the net electric charge is represented by Q (enclosed within the surface), then, we have

$Q=\phi {{\varepsilon }_{0}}$


Therefore, the formula for Gauss’ law is expressed in the terms of net electric charge as

$\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$

where,

  • Q represents the net charge enclosed by a given specific surface, and

  • ${{\varepsilon }_{0}}$ is the electric permittivity constant.

This law explains that the net electric flux through a closed surface depends on the total electric charge contained in the volume within the surface.


Electric Field Due to a Uniformly Charged Infinite Plane Sheet 

Let us consider a charged infinite plane sheet and the charges are uniformly distributed on the sheet. Let P be any  arbitrary point at ‘r’ distance from the sheet. Since, the plane is considered to be infinitely large. Therefore, the electric field at all the points equidistant from the plane sheet would be the same and it would be radially directed at all the points. 


Now, we consider a hypothetical cylindrical surface of length ‘2r’ and area of the plane surface be ‘A’. We assume that the sheet passes through the middle of this surface and is perpendicular to it. This hypothetical closed surface is known as the Gaussian Surface. The above situation is explained in the diagram given below.


Gaussian Surface for Uniformly Charged Infinite Plane Sheet


Gaussian Surface for Uniformly Charged Infinite Plane Sheet


Now, we apply the Gauss’ Law to the hypothetical Gaussian Surface in the above diagram. The net electric flux through the surface will be determined by integrating the product of electric field E and and the area element dA, i.e. 

$\begin{align}& {{\phi }_{E}}=\oint{E\cdot dA} \\ & \Rightarrow {{\phi }_{E}}=\int{E\cdot dA}+\int\limits_{P}{E\cdot dA}+\int\limits_{P'}{E\cdot dA} \\ \end{align}$


Since the electric field is directed normally to the area element for all the points on the curved surface and is directed in the same direction to the area element on the plane surfaces P and P’, we have 

${{\phi }_{E}}=0+\int\limits_{P}{E\cdot dA}+\int\limits_{P'}{E\cdot dA}$

It will be equal to the charged enclosed within the surface divided by the electric constant ${{\varepsilon }_{0}}$ i.e. 

${{\phi }_{E}}=0+\int\limits_{P}{E\cdot dA}+\int\limits_{P'}{E\cdot dA}=\dfrac{Q}{{{\varepsilon }_{0}}}$

The electric field is uniform through the surface, therefore, we take E out of integration. The charge enclosed can be replaced with the product of charge density and total area of the surface. If $\sigma $ denotes the surface charge density and ‘A’ is the total surface area, then we have 

$\begin{align}& 2E\int\limits_{P}{dA=\dfrac{\sigma A}{{{\varepsilon }_{0}}}} \\ & \Rightarrow 2EA=\dfrac{\sigma A}{{{\varepsilon }_{0}}} \\ & \Rightarrow E=\dfrac{\sigma }{2{{\varepsilon }_{0}}} \\ \end{align}$

In vector form, the above equation can be written as 

$E=\dfrac{\sigma }{2{{\varepsilon }_{0}}}\hat{n}$

Here, $\hat{n}$ is the unit vector in the direction perpendicular to the plane. We can observe from the equation that the electric field due a uniformly charged infinite plane sheet is proportional to the surface charge density of the plane sheet and does not depend on the distance ‘r’ from the plane. 


The electric field at any point away from the plane will be the same, since the charge density will remain constant for a uniformly charged plane. From the above equation, we can conclude that if the surface charge density, $\sigma >0$ then the electric field will be directed outwards perpendicular to the plane, and if it is negative, i.e. $\sigma <0$, then the electric field is directed inwards perpendicular to the plane. If the above plane sheet were considered finite, then the equation would be valid only for the area in the middle of the sheet.


Summary

The electric field is defined as electrical force per unit charge. The SI unit of measurement of electric field is Volt/metre. The statement of Gauss’ Law mentions that “The total flux contained within a closed surface equals 1/ε0 times the total electric charge enclosed by the closed surface.” It is formulated as $\phi =\dfrac{Q}{{{\varepsilon }_{0}}}$.


The electric field due to a uniformly charged infinite plane sheet is given by $E=\dfrac{\sigma }{2{{\varepsilon }_{0}}}\hat{n}$ where E is the Electric field, $\sigma $ is the surface charge density and ${{\varepsilon }_{0}}$ is the electric constant. The electric field at any point away from the plane will be the same.

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FAQs on Electric Field due to Infinite Plane for JEE

1. What is the intensity of an electric field inside a conductor?

The intensity of an electric field inside a conductor is always zero. The reason is that the charges that conduct electricity are present only on the surface outside the conductor, due to the result of which the electric field is present only at the external surface of the conductor.


Alternatively, it can be explained with the help of Gauss’ Law. Since the charges lie only on the surface and not inside any conductor, the charge density inside the conductor would be zero. This results in the electric field inside the conductor being zero.

2. What will be the electric field inside a spherical shell?

The answer is zero. As we know that there are no charges inside a conductor, the charges are present only on the outer surface of a conductor. Therefore, if we draw a Gaussian Surface inside the spherical shell, then the Gaussian surface will not enclose any charge. This will result in the surface charge density being zero. Therefore, the electric field will also become zero inside a spherical shell.