Question

The figure shows two thin large identical parallel conducting plates placed in air (dielectric constant = 1) with a slab of dielectric constant K placed between them. The width of the small gap between the conducting plates is d while the thickness of the dielectric slab is t. The conducting plates are given equal and opposite charges due to which the strength of the electric field in air region is E as shown. The area of plates perpendicular to the plane of figure is A and consider it to be large. Neglect the edge effects.(The portion ABCD has the same dimensions perpendicular to the plane of the figure as that of plate or dielectric medium). Mark the correct options :

• A. The net charge in the portion ABCD (volume) is zero
• B. The magnitude of net charge in the portion ABCD is $\frac{{\epsilon }_{0}EA}{K}$
• C. The electrostatic energy stored in the dielectric medium is $\frac{{\epsilon }_0{E}^{2}At}{2K}$
• D. The electrostatic energy stored in the dielectric medium is $\frac{{\epsilon }_0{E}^{2}At}{2K^2}$

Answer 1

Answer: (B), (C)

The magnitude of net charge in the portion ABCD is $\frac{{\epsilon }_{0}EA}{K}$
The electrostatic energy stored in the dielectric medium is $\frac{{\epsilon }_0{E}^{2}At}{2K}$

The electric field inside the conducting plate is zero and electric field inside the dielectric is $\frac{E}{K}$

Now, net flux$=-\frac{E}{K}\times A+0\times A$

Using gauss law,

${\varphi }_{net}=\frac{Q_{net}}{{ϵ}_0}\Rightarrow Q_{net}={ϵ}_0\times (-\frac{EA}{K})$

$|Q|=\frac{EA{ϵ}_0}{K}$

Electric energy density $=\frac{1}{2}{ϵ}_{0}K{E}^2$

Electric energy in dielectric $=\frac{1}{2}{ϵ}_{0}K\times \frac{E^2}{K^2}\times At=\frac{E^2{ϵ}_{0}At}{\left(2K\right)}$