As we know, the electric force per unit charge describes the electric field. The electric field determines the direction of the field. We know that the electric field is directed radially outward for a positive charge, and for a negative point charge, the electric field is directed inwards. Electric fields originate from electric charges or from time-varying magnetic fields. Electric fields are one of the key factors for the electromagnetic force and are one of the four fundamental forces of nature.
In this article, let us learn about 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.
\(\begin{array}{l}Therefore\, electric\, flux\, over\, these\, edges\, \vec{E}.\hat{n}\, ds\, are\, equal\, and\, add\,up.\end{array} \)
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
\(\begin{array}{l}q = \sigma * A\end{array} \)
Hence,
\(\begin{array}{l}2EA = \frac{q}{\epsilon _{0}} = \frac{\sigma A}{\epsilon _{0}}\end{array} \)
or
\(\begin{array}{l}E = \frac{\sigma }{2\epsilon _{0}}\end{array} \)
Vectorially,
\(\begin{array}{l}\vec{E} = \frac{\sigma }{2\epsilon _{0}}\hat{n}\end{array} \)
Where,
\(\begin{array}{l}\hat{n}\, =\, a\, unit\, vector\, normal\, to\, the\, plane\, and\, going\, away\, from\, it\end{array} \)
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.
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.
Read more about Electricity and magnetism
An electric field is defined as the electric force per unit charge and is represented by the alphabet E.
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Consider a thin plane infinite sheet having positive charge density σ. In order to find the electric field intensity at a point p, which is at a perpendicular distance r from the plane shell, we choose a closed cylinder of length 2r, whose ends have an area A as our Gaussian surface. The cylinder is arranged to pierce the sheet perpendicularly as shown.
The total electric flux through the Gaussian surface is the algebraic sum of flux due to cylindrical part and flux due to end caps of the Gaussian surface.Over the cylindrical part, the field lines are parallel to the surface. The angle between
Therefore, the magnitude of the field does not depend on the distance from the plate. This is because as we move farther from the plane sheet of charge, more and more charges comes into our "field of view" and compensates the decrease in field due to increase in distance. For a positively charged infinite sheet, the electric field points normally away from the plane and points normally into the sheet for a negatively charged plane sheet.
Electric field due to two infinite plane parallel sheets of charge:
Consider two infinite plane parallel sheets of charge A and B. Let σ1 and σ2 be the uniform surface charge on A and B respectively.
Let σ1 > σ2. The two sheets of charge divide the space into three regions, region I to the left of sheet A, II between the sheets and III to the right of sheet B
If the two sheets have equal and opposite uniform surface densities of charge i.e., σ1 = σ and σ2 = – σ. Then
Therefore, the electric field between the sheets is constant i.e., a uniform field which is directed from the positive to negative sheet. While the field is zero on the outside of the two sheets. This arrangement is used for producing uniform electric field.
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Let us today again discuss another application of gauss law of electrostatics that is to calculate Electric Field Due To Two Infinite Parallel Charged Sheets:- Consider two parallel sheets of charge A and B with surface density of σ and –σ respectively .The magnitude of intensity of electric field on either side, near a plane sheet of charge having surface charge density σ is given by E=σ/2ε0
And it is directed normally away from the sheet of positive charge.
The resultant electric field intensity E at any point near the sheet,due to both the sheets A and B will be the vector sum due to the individual intensities set up by each sheet (try to make figure yourself).
We will now calculate the intensity of electric field at different points when the surface density of sheet B changes from +σ to –σ.The electric field E2 produced by it will be in opposite direction
At Point P:
At point P ,to the left of the sheets the intensities E1 and E2due to both the sheets are in opposite directions.As they are equal in magnitude,the resultant intensity E would be zero,that is,
E=E1-E2=σ/2ε0 –σ/2ε0=0
At point Q:
At a point Q ,mid way between the sheets,the intensities E1 and E2due to individual sheets are directed normally away from the sheet A or towards the sheet B .therefore, the resultant intensity E at Q is given as
E=E1+E2= σ/2ε0 +σ/2ε0
Or E=σ/ε0
At Point R:
At a point R to the right of sheets,the intensities E1 and E2 are again in opposite directions.Since they are of equal magnitude ,the resultant intensity E would be zero,that is,
E=E1-E2= -σ/2ε0+ σ/2ε0=0
Conclusion. E due to two oppositely charged infinite plates is σ/ε0 at any point between the plates and is zero for all external points.
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