A) 11.2 \[\mu \]f
B) 5.6 \[\mu \]f
C) 4.0 \[\mu \]f
D) 22.4 \[\mu \]f
Correct Answer: A
Solution :
The medium between plates is air, then capacitance is given by \[{{C}_{air}}=\frac{{{\varepsilon }_{0}}A}{d}\] ?(i) \[\left( \begin{align} & \text{Given:}{{d}_{1}}=0.4\,cm \\ & K=2.8 \\ & {{d}_{2}}=\frac{0.4}{2}=0.2\,cm \\ \end{align} \right)\] \[{{C}_{air}}=\frac{{{\varepsilon }_{0}}A}{0.4\times {{10}^{-2}}}\] ?(i) When dielectric is introduced between the plates of parallel plate capacitor \[{{C}_{med}}=\frac{K{{\varepsilon }_{0}}A}{{{d}_{2}}}=\frac{2.8{{\varepsilon }_{0}}A}{0.2\times {{10}^{-2}}}\] ?(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{C}_{med}}}{{{C}_{air}}}=\frac{2.8{{\varepsilon }_{0}}A}{0.2\times {{10}^{-2}}}\times \frac{0.4\times {{10}^{-2}}}{{{\varepsilon }_{0}}A}\] \[{{C}_{med}}=2.8\times 2\times 2=11.2\,\mu F\]
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