Question:

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:

Updated On: Aug 1, 2022
  • $0 \,V/m$
  • $\sigma / 2 \varepsilon_{0} V / m$
  • $\sigma / \varepsilon_{0} V / m$
  • $2 \sigma / \varepsilon_{0} V / m$
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The Correct Option is C

Solution and Explanation

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.
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 $
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Concepts Used:

Gauss Law

Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.

Gauss Law:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.

Gauss Law Formula:

As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;

Q = ϕ ϵ0

The Gauss law formula is expressed by;

ϕ = Q/ϵ0

Where,

Q = total charge within the given surface,

ε0 = the electric constant.