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  4. Two infinitely long parallel conducting plates having surface charge densities +σ and -σ respectively, are separated by a small distance. The medium between the plates is vacuum. It ε0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is:

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Q. Two infinitely long parallel conducting plates having surface charge densities $+\sigma$ and $-\sigma$ respectively, are separated by a small distance. The medium between the plates is vacuum. It $\varepsilon_{0}$ is the dielectric permittivity of vacuum, then the electric field in the region between the plates is:

AIIMSAIIMS 2005Electrostatic Potential and Capacitance

Solution:

Given that conducting plates have surface charge densities $+\sigma$ and $-\sigma$ respectively. Since the sheet is large, the electric field $F$ at energy point near the sheet will be perpendicular to the sheet.
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The resultant electric field is given by
$E'=E+E=2 E$
If $\sigma$ is surface charge density then, electric field
$ E'=\frac{\sigma'}{2 \varepsilon_{0}}$
$\therefore 2 E =\frac{2 \sigma}{2 \varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}} V / m $