A parallel plate capacitor with air between the plates has a capacitance of $8 \;pF \left(1 \;pF =10^{-12} \;F \right) .$ What will be the capacitance (in $pF$) if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6 ?$
A parallel plate capacitor with air between the plates has a capacitance of $8 \;pF \left(1 \;pF =10^{-12} \;F \right) .$ What will be the capacitance (in $pF$) if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6 ?$
Capacitance between the parallel plates of the capacitor, $C =8\, pF$ Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, $k =1$ Capacitance, $C$, is given by the formula,
$C=\frac{k \varepsilon_{0} A}{d}=\frac{\varepsilon_{0} A}{d} \ldots (i)$
Where, $A=$ Area of each plate
$\varepsilon_{0}=$ Permittivity of free space
If distance between the plates is reduced to half, then new distance, $d _{1}= d / 2$
Dielectric constant of the substance filled in between the plates, $k_{1}=6$
Hence, capacitance of the capacitor becomes
$C_{1}=\frac{k_{1} \varepsilon_{0} A}{d_{1}}=\frac{6 \varepsilon_{0} A}{\frac{d}{2}}=\frac{12 \varepsilon_{0} A}{d} \ldots (ii)$
Taking ratios of equations $(i)$ and $(ii)$, we obtain
$C _{1}=2 \times 6\, C =12\, C =12 \times 8 \,pF =96 \,pF$
Therefore, the capacitance between the plates is $96 \,pF$.
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