Question 88 of 180

A parallel plate capacitor with width $4cm$, length $8cm$ and separation between the plates of $4mm$ is connected to a battery of $20V$. A dielectric slab of dielectric constant $5$ having length $1cm$, width $4cm$ and thickness $4mm$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be _____ ${ϵ}_{0}J$. (Where ${ϵ}_{0}$ is the permittivity of free space)

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Answer: 240

The figure below shows the parallel plate capacitor with dielectric slab placed in between the plates.

Now, the effective capacitance of the combination is
${C}_{eff}=\left[\dfrac{{\epsilon }_{0}\left(7\times 4\right)}{\dfrac{4}{10}}+\dfrac{5{\epsilon }_{0}\left(1\times 4\right)}{\dfrac{4}{10}}\right]\times {10}^{-2}$
$=\left[70{\epsilon }_{0}+50{\epsilon }_{0}\right]\times {10}^{-2}=120{\epsilon }_{0}\times {10}^{-2}$
$\Rightarrow {C}_{eff}=1.2{\epsilon }_{0}$
Electrostatics energy of this system is $=\dfrac{1}{2}{C}_{eff}{V}^{2}$
$=\dfrac{1}{2}\left(1.2\right){\epsilon }_{0}\left(20\right)\left(20\right)=240{\epsilon }_{0}$