Answer to Question #200359 in Electrical Engineering for A jaif

Question

A parallel plate capacitor has its plates separated with a slab of 4mm thickness and a dielectric

constant of 3. If the capacitance is to be one-third of the original value when a second slab of

6mm thickness is inserted between the plates, what should be the relative permittivity of the

second slab?

Answer 1

$C_0=\epsilon_1\epsilon_0A/d_1$

When the second plate is installed

$\frac{1}{3}C_0=\frac{C_0C}{C_0+C}\to\frac{1}{3}=\frac{C}{C_0+C}=\frac{\epsilon_2\epsilon_0A/d_2}{\epsilon_1\epsilon_0A/d_1+\epsilon_2\epsilon_0A/d_2}$

$\frac{1}{3}=\frac{\epsilon_2/d_2}{\epsilon_1/d_1+\epsilon_2/d_2}\to \epsilon_2=\frac{\epsilon_1d_2}{2d_1}=\frac{3\cdot 0.006}{2\cdot0.004}=2.25$ . Answer