Question

A parallel plate capacitor with air between the plates has a capacitance of $8\,pF$($1\,pF\,=\,{{10}^{-12}}F$). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?

Answer 1

A capacitor is an energy storing device made of two parallel plates and a dielectric material kept in between the plates and capacitance is the ability of a capacitor to collect and store energy in form of electrical charges. Capacitance is defined as the ratio of charge on capacitor to potential difference.

Complete step by step solution:
Capacitance for a parallel plate capacitor with air or free space in between the plates of the capacitor is given by:
$C\,=\,\dfrac{{{\varepsilon }_{o}}A}{d}$
Also, Capacitance for a parallel plate capacitor with dielectric material in between the plates of the capacitor is given by:
$C=\dfrac{k{{\varepsilon }_{o}}A}{d}$ ……(i)
Here,
Capacitance,$C$ of the capacitor is measured in$F$ (Farad).
Distance between the plates is given by $d$in $m$(meters).
Area of plates is given by $A$in ${{m}^{2}}$(meter sq).
Dielectric constant is given by$k$.
Permittivity of free space is given by${{\varepsilon }_{o}}\,=\,8.84\text{ }\times \text{ }{{10}^{-}}^{12}$in ${F}/{m}\;$(Farad per meter).
Now, according to the question:
Capacitance of parallel plate capacitor is, $C\,=\,8\,pF$ ……(ii)
Dielectric constant of air is,$k\,=\,1$
$\Rightarrow \,C\,=\,\dfrac{{{\varepsilon }_{o}}A}{d}$ ……(iii)
By using equations (ii) and (iii), we get
$\dfrac{{{\varepsilon }_{o}}A}{d}\,=\,8\,pF$ ……(iv)
Now, if the distance between the plates is reduced to half, then the new distance is given by ${{d}^{'}}\,=\,\dfrac{d}{2}$
And, the dielectric constant of the substance filled in between is $k\,=\,6$
Introducing a dielectric material in between the plates of the capacitor will decrease the potential difference between the plates of the capacitor thus, the capacitance increases.
Also, the distance between the plates is inversely proportional to the capacitance. So, decreasing the distance between plates of the capacitor will eventually increase the capacitance.
The area of plates of the capacitor is directly proportional to capacitance. Reducing the area of plates will decrease the capacitance.
Hence, the capacitance of the capacitor becomes:
$\begin{aligned} & \,\,\,\,\,\,C\,=\,\dfrac{k{{\varepsilon }_{o}}A}{{{d}^{'}}} \\\ & \Rightarrow \,C\,=\,\dfrac{6{{\varepsilon }_{o}}A}{{d}/{2}\;} \\\ & \Rightarrow \,C\,=\,\dfrac{12{{\varepsilon }_{o}}A}{d} \\\ \end{aligned}$
Now, by using equation (iv), we get
$\begin{aligned} & \,\,\,\,\,C\,=\,12\times \,8\,pF \\\ & \Rightarrow \,C\,=\,96pF \\\ \end{aligned}$
Therefore, the capacitance of the capacitor with dielectric constant equal to 6 and the distance between plates reduced to half is $96\,pF$ or $96\,\times \,{{10}^{-12}}\,F$.

Note:
Capacitors can be connected in two ways in series combination or in parallel combination. Depending on their connection the capacitance will vary accordingly. When connected in a series combination capacitance decreases as $\dfrac{1}{C}\,=\,\dfrac{1}{{{C}_{1}}}\,+\,\dfrac{1}{{{C}_{2}}}\,+\,...$. When connected in a parallel combination capacitance increases as $C\,\,=\,{{C}_{1}}\,+\,{{C}_{2}}\,+\,...$