Question 8

Two metallic plates form a parallel plate capacitor. The distance between the plate is '$$d$$'. A metal sheet of thickness $$\frac{d}{2}$$ 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?

initially capacitance =

$$\left(\frac{A\varepsilon_0}{d}\right)=C$$

when sheet of thickness $$t=\left(\frac{d}{2}\right)$$ is introduced then capacitance be C'.

and $$C'=\frac{A\varepsilon_0}{d-t}$$

$$\left(\frac{C'}{C}\right)=\frac{\left(d-t\right)}{d}=\frac{\left(d-(d/2)\right)}{d}\ $$

$$\frac{C'}{C}=\frac{2}{1}$$

ratio of capacitance is 2 : 1

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