In a parallel plate capacitor set up, the plate area of capacitor is $2 \,m ^{2}$ and the plates are separated by $1\, m$. If the space between the plates are 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 $=3.2$ ) and (Round off to the Nearest Integer)
JEE MAIN 2021, Diffcult
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$C=\frac{\varepsilon_{0} A }{\frac{ d }{2 K }+\frac{ d }{2}}=\frac{2 \varepsilon_{0} A }{\frac{ d }{ K }+ d }$
$=\frac{2 \times 2 \varepsilon_{0}}{\frac{1}{3.2}+1}=\frac{4 \times 3.2}{4.2} \varepsilon_{0}$
$=3.04 \,\varepsilon_{0}$
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