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Question
Using Gauss's law in electrostatics, deduce an expression for electric field intensity due to a uniformly charged infinite plane sheet. If another identical sheet is placed parallel to it, show that there is no electric field in the region between the two sheets ?
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Solution
Electric Field Due to a uniformly charged infinitely large plane thin sheet with surface charge density σ, using Gauss's law
Consider an infinite thin plane sheet of positive charge with a uniform surface charge density σ on both sides of the sheet. Let P be the point at a distance a from the sheet at which the electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.
The electric flux crossing through the Gaussian surface,
Φ = E × Area of the circular caps of the cylinder
Since electric lines of force are parallel to the curved surface of the cylinder, the flux due to the electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.
Φ = E × 2A … (1)
According to Gauss' Theorem,
`ø q/ε_0`
Here, the charge enclosed by the Gaussian surface,
q = σA
`ø = (σA)/c_0` ...... (2)
From equations (i) and (ii), we get:
`Exx 2A =(σA)/ε_0`
`E = σ/(2ε_0)`
Consider two infinite plane parallel sheets of charge A and B. Let σ1 and σ1 be the uniform surface charge on A and B respectively.
Therefore net electric field (E) between the two sheets can be found out as follows:
\[\text { Electric field due to sheet A }\]
\[ E_1 = \frac{\sigma_1}{2 \epsilon_o} \]
\[\text { Electric field due to sheet B }\]
\[ E_2 = \frac{\sigma_1}{2 \epsilon_o}\]
\[E = E_1 - E_2 \]
\[ = \frac{\sigma_1}{2 \epsilon_o} - \frac{\sigma_1}{2 \epsilon_o} = 0\]
Hence, there is no electric field in the region between the two sheets.
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