Electric Field Intensity Due to a Thin Uniformly Charged Infinite Plane Sheet

Derivation of the electric field intensity due to a thin uniformly charged infinite plane sheet

The uniform surface charge distribution on an infinite plane sheet is represented as σ. In the above figure, the x-axis is normal to the given plane. The electric field does not depend on y and z coordinates, and its direction at each point should be parallel to the x-axis.

The Gaussian surface considered in the above case is a rectangular parallelepiped having a cross-sectional area A.

The unit vector 1 is placed in the -x direction normal to surface 1 and the next unit vector is placed in the +x direction, normal to surface 2.

The total flux through the Gaussian surface = 2 EA.

It is noticed that only faces 1 and 2 will contribute to the flux. The electric field lines are perpendicular to the other faces and do not contribute to the total flux.

In the closed surface, the net charge enclosed is given by

Hence,

or

Vectorially,

Where,

The nature of charge distribution decides the direction of the electric field vector.

If, σ > 0, it conveys that E is directed outwards.

If, σ < 0, it conveys that E is directed inwards.

It is clear from the above expression that E is independent of the distance of the point from the plane charged sheet.

Electric field Intensity Due to Infinite Plane Parallel Sheets

Consider two plane parallel sheets of charge A and B. Let σ¹ and σ² be uniform surface charges on A and B.

Electric field due to sheet A is

\(\begin{array}{l}E_{1} = \frac{\sigma _{1}}{2\epsilon _{0}}\end{array} \)

Electric field due to sheet B is

\(\begin{array}{l}E_{2} = \frac{\sigma _{2}}{2\epsilon _{0}}\end{array} \)

\(\begin{array}{l}= \frac{\sigma _{1}}{2\epsilon _{0}} – \frac{\sigma _{2}}{2\epsilon _{0}} = 0\end{array} \)

Note: No electric field exists in the region between two sheets parallel to each other.

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Frequently Asked Questions on Electric Field Intensity

Q1

1. What is an electric field?

An electric field is defined as the electric force per unit charge and is represented by the alphabet E.

Q2

2. What is the formula to find the electric field intensity due to a thin, uniformly charged infinite plane sheet?

\(\begin{array}{l}E = \frac{\sigma }{2\epsilon _{0}}\hat{n}\end{array} \)

Q3

3. How is the uniform distribution of the surface charge on an infinite plane sheet represented as?

\(\begin{array}{l}\sigma\end{array} \)

Q4

4. What is the net flux through the Gaussian surface?

2 EA.

Q5

5. What is the formula for an electric field?

E = F/q.

Electrical Force and Its Characteristics

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