A parallel plate capacitor has a capacitance of 100 pF , when the plate of the capacitor are separat

Question

A parallel plate capacitor has a capacitance of 100 pF , when the plate of the capacitor are separated by a distance of $t$. Then a metallic foil of thickness $\frac{\mathrm{t}}{3}$ is introduced between the plates. The capacitance will then become\

Professor PC Thomas Classes · Answer

The capacitor works as two capacitors in series and plates of the capacitors are separated $b t / 3$\ \ $\because \mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{t}}, 	herefore \mathrm{C}_{1}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{t} / 3}=3 \mathrm{C}$ andC $_{2}=3 \mathrm{C}$\ $	herefore \mathrm{C}^{\prime \prime}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}=\frac{9}{6} \mathrm{C}=\left(\frac{3}{2}ight) 100 \mathrm{pF}$.\