Question

In a parallel plate capacitor set up, the plate area of the capacitor is 2 $m^2$ and the plates are separated by 1 m. If the space between the plates is filled with a dielectric material of thickness 0.5 m and area 2 $m^2$ (see fig) the capacitance of the set-up will be ________$\varepsilon_0$. (Dielectric constant of the material = 3.2) (Round off to the Nearest Integer)

Answer 1

Hint: In this problem, the capacitance of a parallel plate capacitor must be determined. $C=\dfrac{K{{\in }_{o}}A}{d}$ is the formula for a parallel plate capacitor. The net capacitance for capacitors connected in parallel is calculated using the formula \[{{C}_{net}}={C_1}+{C_2}\]. Additionally, the charge of a parallel plate capacitor is determined by$Q=CV$, and the parallel plate capacitor's electric field is determined by\[E=Vd\].

Complete answer:
The ratio of the magnitude of the charge on each conductor to the magnitude of the potential difference (PD) between them is known as the capacitance (C) of a capacitor.
Hence, we can say that capacitance is equal to the ratio charge is to potential difference.
The value of capacitance is always positive. The coulomb per volt or farad is the S.I. unit for capacitance (F).
The size, shape, relative placements of the plates, and the medium between them all affect the capacitance value. The charge on the plate or the distance between the plates has no bearing on the value of C.
The capacitor's capacitance is determined by $C=\dfrac{{{\varepsilon }_{0}}A}{d}$.

According to the question,
${{C}_{1}}=\dfrac{K{{\varepsilon }_{0}}A}{{d}/{2}\;}$ and ${{C}_{2}}=\dfrac{{{\varepsilon }_{0}}A}{{d}/{2}\;}$
OR
$C=\dfrac{{{E}_{0}}A}{\varepsilon \left( \dfrac{d}{K} \right)}=\dfrac{{{E}_{0}}A}{\dfrac{d}{2K}+\dfrac{d}{2}}$

Now,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$
$\Rightarrow \dfrac{1}{C}=\dfrac{d}{2K{{\varepsilon }_{0}}A}+\dfrac{d}{2{{\varepsilon }_{0}}A}$
$\Rightarrow \dfrac{1}{C}=\dfrac{2{{E}_{0}}AK}{d(K+1)}$
$\Rightarrow \dfrac{1}{C}=\dfrac{d}{2{{\varepsilon }_{0}}A}\left( \dfrac{K+1}{K} \right)$
$C=\dfrac{2{{\varepsilon }_{0}}AK}{d(K+1)}=\dfrac{2\times 2\times 3.2}{1\times 4.2}{{E}_{0}}$
$C=3.04{{E}_{0}}$

Hence, the answer is 3.04.

Note: Sparking occurs when the two capacitor plates are connected by a conducting wire, demonstrating the transformation of electrical energy into heat and light energy. If a medium with a dielectric constant of K is put between a capacitor's plates, the capacitance of the capacitor will increase K times.