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The distance between two plates of a capacitor is d and its capacitance is C_{1}, when air is the medium between
the plates. If a metal sheet of thickness \frac{2d}{3} and of the same area as plate is introduced between the plates, the

capacitance of the capacitor becomes C_{2}. The ratio \frac{C_{2}}{C_{1}} is 

Option: 1

4:1


Option: 2

3:1


Option: 3

2:1


Option: 4

1:1


Answers (1)

best_answer

 

\begin{aligned} & \mathrm{t}=\frac{2 \mathrm{~d}}{3} \\ & \mathrm{~K}=\infty \text { for metals } \\ & \mathrm{C}_1=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \end{aligned}

\mathrm{C}_2=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{k}}}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}-\frac{2 \mathrm{~d}}{3}+0}=\frac{3 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}

=\frac{\mathrm{C}_2}{\mathrm{C}_1}=\frac{\frac{3 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}}{\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}}=\frac{3}{1}

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sudhir.kumar

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