Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $16 \times 10^{-22}\, C\, m^{-2}$. The electric field between the plates is
- $1.8 \times 10^{-10}\, N\, C^{-1}$
- $1.9 \times 10^{-10}\, N\, C^{-1}$
- $1.6 \times 10^{-10}\, N\, C^{-1}$
- $1.5 \times 10^{-10}\, N\, C^{-1}$
The Correct Option is A
Solution and Explanation
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LIST I LIST II A Gauss's Law in Electrostatics I \(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\) B Faraday's Law II \(\oint \vec{B} \cdot d \vec{A}=0\) C Gauss's Law in Magnetism III \(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\) D Ampere-Maxwell Law IV \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) - JEE Main - 2023
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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.