Answer to Question #168558 in Electricity and Magnetism for Vinit Kumawat

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

Let's write the capacitance of the capacitor in the first case when its plates separated with a slab of 4 mm thickness and a dielectric constant of 3:

$C_1=\kappa_1\epsilon_0\dfrac{A}{d_1}.$

Then, a second slab of 6 mm thickness is inserted between the plates:

$C_2=\kappa_2\epsilon_0\dfrac{A}{d_2}.$

Let's divide $C_1$by $C_2$:

$\dfrac{C_1}{C_2}=\dfrac{\kappa_1}{\kappa_2}\dfrac{d_2}{d_1}.$

From this formula we can find the relative permittivity of the second slab:

$\kappa_2=\dfrac{C_2}{C_1}\kappa_1\dfrac{d_2}{d_1},$ $\kappa_2=\dfrac{\dfrac{1}{3}C_1}{C_1}\cdot3\cdot\dfrac{6\ mm}{4\ mm}=1.5.$