Question

A parallel plate capacitor having area $A$ and separated by distance $d$ is filled by a copper plate of thickness $b$. The new capacity is:
A. $\dfrac{{{\varepsilon _ \circ }A}}{{d + \dfrac{b}{2}}}$
B. $\dfrac{{{\varepsilon _ \circ }A}}{{2d}}$
C. $\dfrac{{{\varepsilon _ \circ }A}}{{d - b}}$
D. $\dfrac{{2{\varepsilon _ \circ }A}}{{d + \dfrac{b}{2}}}$

Answer 1

The question involves the use of a parallel plate capacitor. A parallel plate capacitor is an electrical device capable of storing charge across its plate when a potential is applied across its terminals. The amount of charge stored per unit voltage is its capacitance.This quantity depends on a number of factors.Some of which are as follows.
1. Area of the parallel plate capacitor.
2. Distance between the plates.
3. Material between the parallel plates.

Complete step by step solution:
An arrangement of electrodes and insulating material or dielectric is created by parallel plate capacitors. The plates are charged and an electric field is formed between them when two parallel plates are attached via a battery and this configuration is known as the parallel plate capacitor.
A parallel plate, which consists of two metal plates with a distance between them is the simplest configuration for a capacitor. Electrons are deposited on one plate while an equal number of electrons are withdrawn from the other plate.
This very specific property of a parallel plate capacitor makes it ideal for extracting harmonics from the supply of AC. For electronic circuits for different purposes, a parallel plate capacitor is used. It can be used in transducers also.
In a parallel plate capacitor, the electric field $E$ is directly proportional to the charge $Q$ as there will be more field lines if there is more charge.
So,
$\Rightarrow E\;\propto\; Q$ ...... (i)
Also, voltage across the parallel plates is given by:
$\Rightarrow V = Ed$
Thus,
$\Rightarrow V\;\propto\; E$ ...... (ii)
From equation (i) and (ii), we get:
$\Rightarrow Q\;\propto \;CV$
Different capacitors hold different quantities of charge for the same applied voltage, based on their physical characteristics. We define their capacitance $C$ to be such that the load $Q$ contained in the capacitor is proportional to the load. The charge deposited in the capacitor is given by:
$\Rightarrow Q = CV \\\ \Rightarrow C = \dfrac{Q} {V}$
At the point when a voltage $V$ is applied to the capacitor, it stores a charge $Q$ , as appeared. We can perceive how its capacitance relies upon surface area $A$ and distance d by considering the qualities of the Coulomb power. We realize that like charges repulse, not at all like charges pull in, and the force between charges falls down because of separation. So it appears to be very sensible that the greater the plates are, the more charge they can store—on the grounds that the charges can spread out additional. Hence $C$ ought to be more prominent for bigger $A$ Also, the closer the plates are together, the more noteworthy the attraction of the opposite charges on them. So $C$ should be greater for smaller $d$ .
Hence, the generalised equation for a parallel plate capacitor is given as:
$\Rightarrow C = {\varepsilon _ \circ }\left( {\dfrac{A}{d}} \right)$ where ${\varepsilon _ \circ }$ represents the absolute permittivity of the dielectric material being used.
Let the copper plate be filled with thickness $t$
We know that,
$C = {\varepsilon _ \circ }\left( {\dfrac{A} {d}} \right) \\\ \Rightarrow C' = \dfrac{{A{\varepsilon _ \circ }}} {{\left( {(d - t) + \dfrac{t} {k}} \right)}} \\\$
Where $A$ is the area of the capacitor
$d$ is the distance between the plates of capacitor
$t$ is the thickness of dielectric inserted between the plates
$k$ is the dielectric constant and depends on the material
Taking $t = b$ and $k = \infty$
We get
$\Rightarrow C' = \dfrac{{A{\varepsilon _ \circ }}}{{\left( {d - b} \right)}}$
The new capacity is $C' = \dfrac{{A{\varepsilon _ \circ }}}{{\left( {d - b} \right)}}$

Hence, option C is correct.

Note:
The effect of inserting a material of differing electric permittivity causes the change in the capacitance value of the parallel plate capacitor. A number of dielectric materials of differing thicknesses can be inserted between the plates.The capacitance can be calculated in a similar way.