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Two metallic plates form a parallel plate capacitor. The distance between the plate is \mathrm{'d'}. A metal sheet of thickness \mathrm{\frac{d}{2}} and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?

Option: 1

\mathrm{2:1}


Option: 2

\mathrm{1:2}


Option: 3

\mathrm{1:4}


Option: 4

\mathrm{4:1}


Answers (1)

best_answer

Let the new and original capacitance be \mathrm{C\: and \: C_{0}} respectively

  \mathrm{C_{0}=\frac{A \epsilon _{0}}{d} }\\

\mathrm{C=\frac{A\epsilon_{0}}{\frac{t_{1}}{k_{1}}+\frac{t_{2}}{k_{2}}}}

     \mathrm{=\frac{A \epsilon_{0}}{\frac{d / 2}{\infty}+\frac{d / 2}{1}}}\\                  ( Dielectric constant of metal is \mathrm{\infty} )

\mathrm{C=\frac{2 A \epsilon_{0}}{d}=2 C_{0}}\\

\mathrm{\frac{C}{C_{0}}=\frac{2}{1}}

Hence the correct answer is option 1.

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